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Full text of "Two-stage genome search design in affected-sib-pair method"

TWO-STAGE GENOME SEARCH DESIGN IN AFFECTED-SIB-PAIR METHOD 



By 
CHI-HSE TENG 



A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL 

OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT 

OF THE REQUIREMENTS FOR THE DEGREE OF 

DOCTOR OF PHILOSOPHY 

UNIVERSITY OF FLORIDA 

1997 



: 



© Copyright 1997 

by 
CHI-HSE TENG 






ACKNOWLEDGMENTS 

I would like to thank the chairman of my committee, Professor Mark C.K. Yang, 
for introducing me to the field of genetic statistics and his wise guidance throughout 
my graduate study at UF. I also want to thank Professor Randy Carter, Professor 
Frank Martin, Professor Sue McGorrary, and Professor Jin Xiong She. Their advice, 
encouragement, and patience were essential. Last but not least, I want to thank 
Ms. Margaret Joyner for her gracious editorial help and effort. Finally, my deepest 
appreciation goes to my parents, brother, sister, and parents-in-law for their love and 
never-ending support and to my wife Hsiao- Yun and our daughter Gillian for their 
sweet company. 



in 



TABLE OF CONTENTS 



ACKNOWLEDGMENTS iii 

LIST OF TABLES vi 

ABSTRACT viii 

CHAPTERS 

1 INTRODUCTION 1 

2 LITERATURE REVIEW 8 

2.1 Map Function 8 

2.2 An Introduction to Human Linkage Data for Genetic Dissection 11 

2.3 Identical by State, Identical by Descent 16 

2.4 Distributions of IBD 17 

2.5 Risk Ratio 24 

2.6 Test Statistics Based on IBD Score 25 

2.7 Likelihood Ratio Test 30 

2.8 Heterogeneity and Homogeneity 33 

2.9 Polygenes 37 

2.10 Two-stage Genome Search 38 

3 TWO-STAGE GENOME SEARCH FOR SIMPLE MENDELIAN DIS- 

EASE 40 

3.1 Assumptions 41 

3.2 Two-Stage Procedure 42 

3.3 Probability of Allocating the Correct Marker 43 

3.4 Type I Error and Power of Claiming Linkage 72 

3.5 Discussion 76 

4 TWO-STAGE GENOME SEARCH FOR COMPLEX DISEASE .... 78 
4.1 Genetic Model and Assumptions 78 



IV 



4.2 Two-stage Genome Search 79 

4.3 Probability of Allocating the Correct Marker for a Complex 

Disease 81 

4.4 Discussion 110 

5 CONCLUDING REMARKS 112 



REFERENCES 114 

BIOGRAPHICAL SKETCH 120 



LIST OF TABLES 



Table Page 

1.1 Results of Mendel's crosses experiments 2 

1.2 The results of Bateson and Bunnet's experiment 5 

2.1 Multilocus-feasibility of several map functions 12 

2.2 Some sampling issues in linkage studies 16 

2.3 The values of Pr(IBD of trait gene | IBD of marker and relationship). 20 

2.4 The probability of IBD score for different relatives 26 

3.1 The joint distribution of IBD score of the locus 1 and locus 2 57 

3.2 The joint distribution of X,j and X,/j 58 

3.3 Optimal resource allocation in two-stage genome search for rare reces- 
sive gene 64 

3.4 Optimal resource allocation in two-stage genome search for rare dom- 
inant gene 68 

3.5 The 95% percentile of the unique maximum of R Binomial(n 2 , 0.25). . 74 

3.6 The 95% percentile of the unique maximum marker group of R Binomial(n 2 , 
0.25) 75 

4.1 Trait IBD distribution conditional on parental genotype and E vector. 83 

4.2 Exclusive events for the case "Gi is found but G 2 is not." 84 

4.3 Exclusive events for the case "G 2 is found but G\ is no." 85 

4.4 Exclusive events for the case U G\ and G 2 are found." 85 

4.5 Conditional distribution of X\ 3 and X%j given 7\ 87 



VI 



4.6 P(both affected | Ei, E 2 ) and possible trait IBD given E x and E 2 89 

4.7 Optimal resource allocation in two-stage genome search for rare reces- 
sive gene, assumed two genes 95 



vii 



Abstract of Dissertation Presented to the Graduate School 

of the University of Florida in Partial Fulfillment 

of the Requirements for the Degree of 

Doctor of Philosophy 

TWO-STAGE GENOME SEARCH DESIGN IN AFFECTED-SIB-PAIR METHOD 

By 
CHI-HSE TENG 
December, 1997 

Chairman: Mark C.-K. Yang 
Major Department: Statistics 

Since Penrose proposed the sib-pair method in 1935 and the affected-sib-pair 
(ASP) method in 1950 and 1953, these methods have been widely used in linkage 
studies. Combined with contemporary DNA-level genetic markers technology, ASP 
method can now be used for genome- wide searches for disease genes. 

A two-stage search involves first a screening search to eliminate nonviable marker 
loci, then an intensive search to identify gene location using the remaining markers. 
Although this approach has been suggested in the literature, its properties have not 
been thoroughly investigated. The spacing between markers, the number of ASPs to 
be used in the experiment of each stage, and the criteria for markers to pass the first 
stage need to be determined. The major difficulties of the two-stage approach are (1) 
the joint distribution of the nonindependent statistics is difficult to handle; (2) the 
"random" dependence structure of the test statistics in the second stage; and (3) in 
most practically available sample sizes the high number of ties in test scores make 
asymptotic approaches inappropriate. This paper intends to provide a solution for 
designing an optimal design for rare autosomal recessive and dominant diseases. 



vin 



Power computations using the multinomial distribution supported by simulation 
show that in most cases a two-stage design is usually better than a one stage design, 
but not always. Combining data from two stages will gain a small increase in accuracy. 
The optimal designs for resource allocation in each of the two stages are obtained and 
presented in tables. 



IX 



CHAPTER 1 
INTRODUCTION 



Modern genetics began with the work of Gregor Mendel (1822-84) conducted 
between 1856 and 1863, forming the basis for his 1866 paper (Klug and Cummings, 
1997). Gregor Mendel, an Austrian monk, performed a series of experiments on the 
garden peas. Based on the results of these experiments, he proposed a particulate 
inheritance theory which hypothesized that heritable biological characteristics were 
carried and controlled by individual "units." We call these units genes today, but 
Mendel called each unit a "Merkmal," the German word for "Character" (Levine and 
Miller, 1991). 

In Mendel's garden pea experiments, he studied seven characters of garden peas. 
They are: round or wrinkled ripe seeds, yellow or green seed interiors, purple or white 
petals, inflated or pinched ripe pods, green or yellow unripe pods, axial or terminal 
flowers, and long or short stems (Griffiths et al, 1993). For each one of these seven 
characters, Mendel obtained pure lines of plants. "A Pure line is a population that 
all offspring produced by selfing or crossing within the population show the same form 
of the character being studied" (Griffiths et al., 1993). The parental generation, 
denoted as F , in Mendel's experiment are the plants of these pure lines. Mendel first 
studied these characters separately; he crossed two pure lines, one for each phenotype, 
of every characters to obtain the first filial generation, denoted as Fi. All the 
individuals of F! have only one phenotype of each characters. Next, Mendel self- 
fertilized Fi and obtained second filial generation, denoted as F2. The results are 
showed in Table 1.1. Mendel established two principles to explain the pattern of the 



Table 1.1. Results of Mendel's crosses experiments. 



Parental phenotype 


F, 


F 2 


F2 Ratio 


Round x wrinkled seeds 


All round 


5474 round; 1850 wrinkled 


2.96:1 


Yellow x green seeds 


All yellow 


6022 yellow; 2001 green 


3.01:1 


Purple x white petals 


All purple 


705 purple; 224 white 


3.15:1 


Inflated x pinched pods 


All inflated 


882 inflated; 299 pinched 


2.95:1 


Green x yellow pods 


All green 


428 green; 152 yellow 


2.82:1 


Axial x terminal flowers 


All axial 


651 axial; 207 terminal 


3.14:1 


Long x short stems 


All long 


787 long; 277 short 


2.84:1 



Source: Griffiths et al., 1993, p. 23. 

data (Levine and Miller, 1991; Griffiths et al, 1993): 

• The characteristics of an organism are determined by individual units of heredity 
called genes. Each adult organism has two alleles for each gene, one from 
each parent. These alleles are segregated (separated) from each other when 
reproductive cells are formed, each gametes receive one of the two alleles with 
equal chance. This is the principle of segregation, also being known as the 
Mendel's First Law. 



• In an organism with contrasting alleles for the same gene, one allele may be 
dominant over another. This is known as the principle of dominance. 

He assumed that each plant had two alleles for each trait his studied. This assumption 
was correct. "A Allele is one of the different forms of a gene that can exist at a 
single locus" (Griffiths et al., 1993, p. 783). He introduced the notation A for the 
dominant allele and a for the recessive allele. Since the parental generations were 
pure lines, their genotypes were homozygous AA and aa. Thus, the F x population 
were all heterozygous Aa phenotype. The F 2 , offspring of the Fi, were expected to 



be A A, Aa, and aa in the ratio 1:2:1; and dominant (AA and Aa) and recessive 
(aa) phenotype in ratio 3:1. Mendel cultivated 10 seeds from each of 100 dominant F 2 
plants, if all 10 offspring from a single F 2 plant were of the dominant character then he 
concluded this plant was homozygous. Mendel used this experiment to demonstrate 
the 1:2 ratio of homozygous (AA) dominant and heterozygous dominant (Aa) F2's. 
However, for a heterozygous F 2 parent there is still a (0.75) 10 = 0.0563 probability 
would produce 10 dominant offspring, and hence been misclassified as homozygous. 
Therefore, the true expected ratio should be 0.3709:0.6291 instead of 1:2. Fisher 
suggested that Mendel's data are too close to 1:2 rather than the correct values, and 
thus suspected there was some manipulation, or omission of data (Weir, 1996). 

Mendel also studied the two characters together; for example the shape of the 
seeds and the color of the seed. We denote round allele as R and wrinkled allele as r, 
and yellow allele as Y and green allele as y. He crossed a parental line of plants with 
yellow and round seeds (genotype RRYY) with a line with green and wrinkled seeds 
(genotype rryy). The seeds of Fi were all round and yellow (RrYy) showing that 
these two alleles, R and Y, are dominant over r and y. Then, he crossed Fi and got 
four types of seeds, 315 round/yellow seeds, 108 round/green, 101 wrinkled/yellow, 
and 32 wrinkled/green. The ratio was approximately 9:3:3:1 (Levine and Miller, 
1991). All possible genotype combinations of F 2 are given in the following Punnett 
Square (Griffiths et al., 1993); the first row represents the genotype of gamete from 
father, and the first column represents the genotype of gamete from mother, and the 
numbers in the parenthesis are the expected frequencies if the alleles of these two 
characters are segregated independently. 





RY (J) 


Ry(i) 


iy(i) 


rY(J) 


RY (J) 


RRYY (i) 


RRYy (i) 


RrYy (4) 


RrYY (^) 


Ry(i) 


RRYy (A) 


RRyy (A) 


Rryy (4) 


RrYy (4) 


ry(i) 


RrYy (i) 


Rryy (4) 


rr yy (4) 


"Yy (^) 


rY(i) 


RxYY (4) 


RrYy (4) 


"Yy (4) 


"YY (4) 



Mendel established the following principle which could explain the pattern of the 
data. 

• Each of the seven genes that Mendel investigated segregated independently 
(independent assortment), also being known as the Mendel's Second Law 
(Levine and Miller, 1991; McPeek, 1997). 

Regardless of the controversy of his data, Mendel's theory regarding genes controlling 
biological characters is well accepted. 

Correns (1900) observed the phenomenon of complete linkage in which alleles 
of two or more different characters appeared to be always inherited together, rather 
than independently as Mendel's Second Law (McPeek, 1997). This violation led to 
an extension of Mendel's theory; the chromosome theory of heredity which says 
that the genes are parts of specific cellular structures, the chromosomes (Griffiths et 
al., 1993). "In 1902, Walter Sutton and Theodor Boveri, independently published 
papers linking their discoveries of the behavior of chromosomes during meiosis to the 
Mendel's principles of segregation and independent assortment. Sutton and Boveri 
are credited with initiating the chromosomal theory of heredity" (Klug, 1997, p. 61). 
This theory provided a physical mechanism for Mendel's Laws. It were assumed that 
those characters that Mendel studied lay on different chromosomes, and that those 
which were completely linked lay on the same chromosome (McPeek, 1997). 

Bateson and Punnett did an experiment on the sweet peas to study two characters: 
the flower color, purple (dominant) and red (recessive), and the form of pollen, long 



(dominant) and round (recessive). First, they crossed purple, long (PPLL) with 
red, round (ppll) to create a heterozygotes (PpLl) type (Fl generation), then they 
crossed Fl generation among themselves and produced 381 plants. Result is as follows 
(Levine, 1991, p. 193). These results did not match Mendel's Law, nor these two 

Table 1.2. The results of Bateson and Bunnet's experiment. 



Phenotype 


Number 


Percent 


Percent expected 


Percent expected 




observed 


observed 


if unlinked 


if completely linked 


Purple, long 


284 


74% 


56% 


75% 


Purple, round 


21 


6% 


19% 


- 


Red, long 


21 


6% 


19% 


- 


Red, round 


55 


14% 


6% 


25% 



Source: Levine, 1991. 



genes were completely linked. Morgan (1911) provided an explanation for Bateson 
and Punnett's observation and for his own fruit fly experiment's results which is 
similar to Bateson and Punnett's. He suggested the exchange of genetic material had 
occurred between two homologous chromosomes when they paired during cell division, 
and this exchange was called a crossover (McPeek, 1997). Recombination is the 
process by which progeny derive a combination of genes different from that of either 
parent (DOE, 1992). Each individual inherits one chromosome from its father and 
the other one from its mother. When this individual reproduces two chromosomes to 
pass to its offspring during meiosis, it will not pass one of each pair that it inherited 
from its parents but a "blended" one. Meiosis produces crossovers that blend two 
chromosomes and cause recombinations. If there is no crossover or an even number 
of crossovers between two loci, then recombination does not happen to these two 
loci. On the other hand, if there are odd number of crossovers, then recombination 
happens. The probability of recombination happening between two loci is called 



the recombination fraction. At the state of current technology level, cross-overs 
are not directly observable but recombination between two genes (or markers) are 
(Ott, 1991). An obvious consequence of the crossover process, is that the closer two 
genes are to each other on a chromosome, the less chance that crossovers can happen 
between them. Hence there is less chance that recombination can happen between 
them. This idea became the foundation of linkage analysis. Recombination is the 
chief source of variation among species (Rhodes et al., 1974). 

The purpose of linkage study is to find the relative location between genes or 
markers and finally to present them in a linkage map. The distance between two loci 
on a chromosome is measured using genetic distance. We will discuss this in §2.1 
"Map Function". 

Lander and Schork (1994) summarized that there are four major categories of 
methods for genetic dissection. One is "genetic analysis of large crosses in model 
organisms such as the mouse and rat." The other three use human genetic data. These 
three methods are, pedigree analysis, allele-sharing methods, and association studies 
in human population. We will discuss more details in §2.2 "An Introduction to Human 
Linkage Data for Genetic Dissection". Affected-sib-pair method, proposed by Penrose 
(1953), is one of allele-sharing methods. Affected-sib-pair method combined with 
modern DNA-level genetic marker technology, like the Restriction Fragment Length 
Polymorphism markers (RFLP) (Botstein et al., 1980), can be used for genome-wide 
gene search. 

This dissertation will discuss a two-stage genome-wide approach in affected-sib- 
pair method for searching for genetic disease genes. Only the designs for searching 
for rare recessive and dominant diseases with dichotomous phenotype (affected or not 
affected) is studied. Chapter 2 provides a literature review of related topics, includ- 
ing map function, sib-pair method, affected-sib-pair methods, identical by descent, 
identical by state, affected-relative-members method, heterogeneity, and polygeneity. 



Chapter 3 discusses how to design an optimal two-stage search for a simple Mendelian 
disease gene and Chapter 4 discusses how to design an optimal two-stage search for 
two low penetrance, no interaction, unlinked recessive, complex disease genes. Brief 
concluding remarks are given in the last chapter. 



CHAPTER 2 
LITERATURE REVIEW 



In this chapter, the following topics will be covered: 

• map function (without biological interference), 

• concepts of identical-by-descent (IBD) and identical-by-state (IBS), 

• test statistics; based on IBD, IBS scores and likelihood ratio, 

• sib-pair method and affected-sib-pair method, 

• heterogeneity and homogeneity, 

• polygeneity, 

• two-stage approach. 

2.1 Map Function 

The genetic map distance, in units called Morgans, between two genes is defined 
as the expected number of crossovers occurring between two genes on a single chro- 
mosome strand (Ott, 1991). When the distance is so small that the probability of 
multiple crossovers is negligible, the genetic map distance is equivalent to the recom- 
bination fraction. A centimorgan between two genes is a distance that produces an 
approximate 0.01 probability that recombination taking place between them. When 
the distance is large, the probability of multiple crossovers increases. If there is zero 
or an even number of crossovers, then we will not observe recombination, but if there 

8 



is an odd number of crossovers we will observe recombination. Hence, the recombi- 
nation fraction is not necessarily equal to the map distance when two loci are farther 
apart. A map function is used to relate the additive, but hard to measure, ge- 
netic or map distance to the non-additive, but more readily estimable, recombination 
fraction (Speed, 1997). Map function should preserve the additivity of map distance 
but not every map function does. For example, consider three loci, x x , x 2 , x 3 on a 
chromosome and let On and # 23 be the map distance between x\ and x 2 , and x 2 and 
x 3 respectively, and r i2 and r 23 be the recombination fraction distance between x\ 
and x 2 , and x 2 and x 3 respectively. If we assume that x l5 x 2 , and x 3 are very close 
such that the chance of multiple crossovers between them is negligible then the Mor- 
gan map function will not preserve additive of distance but the Haldane (1911) map 
function will. For Morgan map function, r i2 = #12 and r 23 = 23 , the map distance 
between X\ and x 3 , 13 is equal to 12 + # 23 , hence recombination fraction between 
xx and x 3 , ri 3 is equal to r 12 + r 23 . On the other hand, since the chance of multiple 
crossovers is negligible, then 

Pr(recombination happen between x\ and x 3 ) 
= Pr(recombination happen between X\ and x 2 but does not between x 2 and x 3 ) 
+ .Pr(recombination does not happen between Xi and x 2 but between x 2 and x 3 ) 

= 012(1 — #23) + (1 — #12)#23 
= #12 + #23 "~ 2#12#23 

7^ T\% + r 23 . 

This is an contradiction. For Haldane map function, since the map distance is #i 2 +# 23 , 
r l3 is equal to 0.5(1 - exp -2( * 12+(?23 >), and 

Pr(recombination happen between Xi and x 3 ) 
= 0.5(1 - exp- 2 " 12 )[l - 0.5(1 - exp- 2 " 23 )] + 0.5(1 - exp- 2 *")[l - 0.5(1 - exp" 2 ^)] 



10 

= 0.5{(1 -exp -2 * 12 ) + (1 -exp" 2 " 23 ) - (1 - exp- 2 * 12 )(l - exp -2 * 23 )} 
= 0.5(1 -exp- 2(Sl2+e23) ). 

The additive of map distance is preserved. 

Also, "most map functions are determined by three loci comparisons which may 
not be consistent in terms of four or more loci" (Liberman and Karlin, 1984). If a 
map function is consistent for any number of loci then it is called multilocus-feasible. 
Several counterexamples in Ott (1991, p. 127) showed that a non-multilocus-feasible 
might result a negative probability. 

Liberman and Karlin (1984) investigated these problems. Their conclusions can be 
summarized as follows: there are two methods for constructing a genetic map function, 
the first starts with a model of the recombination process, the second uses a differential 
equation method. "The renewal crossover formation process model assumes that 
crossovers occur in succession along the chromosome, starting at a natural biological 
site (e.g., the centromere, some origin of replication, or a telomere) such that the 
lengths of the intervals between successive crossovers are positive random variables 
independently and identically distributed." 

For the first method, if a renewal crossover formation model is specified, the only 
feasible map function is the Haldane map function (Haldane, 1919). The correspond- 
ing crossover formation process is a Poisson process. 

Theorem 1 Consider a crossover formation model of the form of a renewal process 
with intercrossover distribution F(t), the distribution function of the distance between 

two successive crossovers, such that ^r(O) / for some positive integer n. Then a 
map function exists if and only if the intercrossover distribution agrees with that of 
an exponential distribution, i.e., the renewal process is a Poisson process (Liberman 
and Karlin, 1984). 



11 

On the issue of multilocus feasibility of map functions, they derived necessary and 
sufficient conditions for multilocus-feasibility. 

Theorem 2 Let r be recombination fraction and x be genetic distance. Sufficient 
conditions: A map function r = M(x) is multilocus-feasible if its derivative functions 
M* n ) obey the inequalities 

(-l) n M (n) (x) <0, n = 1,2,..., for all x > 0. 

Necessary conditions: Let r = M(x) be a map function that is multilocus-feasible 
and suppose that all derivatives M^ n \x) exist. Then 

(-l)"M' n l(0) <0, Vn=l,2,.... 

Table 2.1 shows the multilocus feasibility of several map functions. Note that the 
multilocus-feasible map functions are constructed based on the cross-over formation 
process method, where r is the recombination fraction and x is genetic map distance. 
Because further extension will incorporate genetical interference, we will not discuss 
it here, but it can be referred to Karlin and Liberman (1994), and Speed (1997). 
Most map functions constructed by a differential equation method are not multilocus 
feasible, also will not be discussed. 

2.2 An Introduction to Human Linkage Data for Genetic Dissection 

The purpose of linkage study is to locate the gene which we are interested in. 
With contemporary DNA-level genetic marker technology (Botstein et al., 1980), one 
way to achieve this is by studying the linkage between genes and markers. "When two 
genes are inherited independently of each other, recombinants and nonrecombinants 
are expected in equal proportions among the offspring. For some pairs of genes, one 
observes a consistent deviation from the 1:1 ratio of recombinant to nonrecombinant 



12 



Table 2.1. Multilocus-feasibility of several map functions. 



Source 


Map Function r = M(x) 


Multilous 
Feasible 


Haldane (1919) 


1(1 -exp- 2 *) 


Yes 


Ludwig (1934) 


|sin(2z) 


No 


Kosambi (1944) 


\ tanh(2x) 


No 


Carter and Falconer 


x = J| tan _1 (2r) + tanh _1 (2r) | 


No 


(1951) 






Sturt (1976) 


|[l-(l-(a:/L))exp-^ 2L - 1 )/ L ] 


Yes 


Rao et al. (1977) 


x=i[ p(2p-l)(l-4p)ln(l-2r) 


No 




+ 16p(l - p)(2p - 1) tan _1 (2r) 






+2p(l - p)(8p + 2) tanh _1 (2r) 
+6(l-p)(l-2p)(l-4p)r] 




Felsenstein (1979) 


l-exp 2 ^- 2 )* 


No 


2(l-(A'-l)exp 2 (' c - 2 ) 1 ) 



Source: Liberman and Karlin, 1984. 



13 

offspring ... In other words, alleles of different genes appear to be genetically 
coupled, and this phenomenon is called genetic linkage." (Ott, 1991, p. 6) A 
marker is an small identifiable physical region on a chromosome and the inheritance 
of this region can be monitored (DOE, 1992). The closer a marker is to the gene, the 
smaller the chance that they will be recombined during the DNA replication process. 
Therefore, if a marker has a strong correlation with the phenotype of a gene, the gene 
should be in proximity to the marker. There are three major categories of methods 
using human genetic data: pedigree analysis, allele-sharing method, and association 
study. Risch and Merikangas (1996) summarized these methods as follows: 

In pedigree analysis, we first collect pedigrees that contain affected members. Next 
we propose a genetic model (location of the gene, allele frequency, mode of inheritance, 
and so on), say Mi, to explain the pattern observed in the data, and compare the 
likelihood under Mi with the likelihood under null hypothesis M , which assumes no 
gene in the region of the marker, by the likelihood ratio, 

L(data\M ) 
L{data\M x y 

or equivalently by logarithm of the odds (LOD) (Barnard, 1949) score, 

L(data\M l ) 



log 



10 



L(data\M )' 



Since this method requires specifying a genetic model, it is mostly used to identify 
simple Mendelian trait genes. 

In allele-sharing methods, we first collect relative pairs or groups (most of the 
time, we collect affected relatives), and try to prove that the pattern observed in 
the data is not due to random Mendelian segregation. Most of these methods use 
identical-by-decent scores, which is the number of identical copies of the markers that 
relatives share. The major advantage is that no specific genetic model is required to 






14 

do the test, so they are considered as nonparametric methods. These methods are 
mostly used for detecting non-Mendelian complex trait genes. 

Association study is a case-control study. Instead of using familial data, rather it 
collects unrelated affected and unaffected individuals, and then compares the allele 
frequencies of the candidate genes or markers. If the disease gene(s) is/are one of 
those candidate gene or very strongly associated with some marker alleles, then the 
affected group will have higher frequencies of those alleles than the control group. This 
method is mostly used after the possible region of the genes has been narrowed down, 
because this method involves examining genes and/or markers on a very fine scale, 
i.e., examining many markers in a small area of chromosome. "Association studies 
seem to be of greater power than linkage studies. But of course, the limitation of 
association studies is that the actual gene or genes involved in the disease must be 
tentatively identified before the test can be performed. . . . Thus, the primary 
limitation of genome- wide association tests is not a statistical one but a technological 
one" (Risch and Merikangas, 1996, p. 1517). 

Allele-sharing methods include the affected-sib-pair (ASP) method and its vari- 
ants. Penrose was considered the first person to propose the sib-pair method (Con- 
neally and Rivas, 1980; Shah and Green, 1994). In 1935, Penrose proposed a simple 
X 2 test, to detect linkage between two characters. In 1938, he extended this method 
to "graded" characters, which means the characters could have intermediate value. 
In 1950, Penrose proposed the concept of affected-sib-pair method as a means for 
analyzing red hair phenotype and ABO blood group data. He wrote: "Accuracy is 
preserved and uninformative dead wood excluded if a set of sibships is selected by the 
presence of one of the test characters . . ."A general form of the sib-pair method was 
proposed by Penrose in 1953. Now, these methods have been extended to examine 
identical- by-descent scores (elaborated in the following sections), identical-by-state 



15 

scores of markers, and the risk ratio between two relatives. The idea of affected-sib- 
pair method is only allow a pair of sibs both who are affected to be included in the 
study. If the disease is caused by a gene then both affected sibs tend to receive the 
same gene. Hence, if a marker is close to the gene, then the marker will tend to 
segregate with the gene during meiosis. We will cover how to detect the linkage in 
the later sections. 

The major advantages of affected-sib-pair method over pedigree analysis were 
summarized by Holmans and Craddock in 1997: 

• Affected-sib-pair method does not require specification of a genetic model, which 
is important for complex disease where the mode of inheritance is unclear. 

• It is generally easier to collect affected sibling data than to collect a large, 
multigeneration pedigree with multiple affected members. 

• Affected sib-pairs are more likely to be informative for linkage than large pedi- 
gree under oligogentic epistatic models, which is plausible for a number of com- 
plex traits. 

The disadvantage of affected-sib-pair method is that it is considered less powerful than 
traditional pedigree analysis when the genetic model can be specified. If the genetic 
model cannot be specified in the affected-sib-pair methods, the recombination fraction 
cannot be estimated. 

Suarez, Rich, and Reich (1978), and Blackwelder and Elston (1985) suggested that 
sampling only affected sib-pairs is more powerful then sampling affected-unaffected 
sibs under a fixed sample size constraint. Some sampling issues in linkage studies as 
summarized by Gershon et al. (1994) appear in Table 2.2. 



16 






Table 2.2. Some sampling issues in linkage studies. 



Issue 


Pro 


Con 


Very large pedigree 

Medium-size 
pedigree series 

Nuclear families 
Affected sib-pairs 


Statistical power high under 
homogeneity assumption. 

Less likely to have het- 
erogeneity within pedigrees. 
More likely to get generaliz- 

able results. 
Model-free 


Extension may actually in- 
troduce heterogeneity. Very 

hard to find. 

Heterogeneity between pedi- 
grees. 

Lower power: very 
many needed if heterogene- 
ity present. 

Even lower statistical power. 



Source: Gershon et al., 1994. 



2.3 Identical by State, Identical by Descent 



IBS stands for "identical by state." It means two alleles are the same regardless 
of whether the alleles are copies of the same ancestral alleles or not. IBD stands 
for "identical by descent." Two alleles are said to be IBD if those two alleles are 
copies of the same ancestral alleles. Since humans are diploid, that is, having two 
haploid sets of chromosomes, two sibs can share 2, 1, or marker alleles at a locus. 
Consequently, their IBD scores would be equal to 2, 1, or 0, respectively. We will 
discuss the distribution of IBD score in the next section. We only consider two alleles 
are identical-by-decent if they are copies of the same ancestral allele in the pedigree 
under study, i.e., if the alleles are traceable. 



17 

With the affected-sib-pair method, since the sib-pairs were selected for study on 
the condition that both sibs are sick with the same disease (biological character), 
if the disease is caused by a gene, then a marker adjacent to the disease gene has 
a higher probability to segregate with the gene during meiosis. Hence, this marker 
tends to have a higher IBD score. This method can serve as a tool to locate the trait 
genes. 

2.4 Distributions of IBD 

Under ideal conditions, in using the affected-sib-pair method, we can take IBD 
scores to make inferences about the recombination fraction between the trait gene(s) 
and the marker (s). Hence it is important to know the distribution of the IBD scores. 
According to Mendel's First Law, each of a parent's haplotypes have an equal chance 
to be passed to offspring. Therefore, if sibs were not selected conditionally on any 
other biological characters, then, the probability for a pair of full sibs to have a IBD 
score equal to 2 is 0.25, to 1 is 0.5, and to is 0.25. If sibs are selected conditional on 
a biological character and if the marker(s) we examine is unlinked to the trait gene(s) 
that responsible for the biological character, then, regardless of the inheritance mode, 
penetrance, and allele frequency of the trait gene the probability mass function is also 
0.25, 0.5, and 0.25 for IBD equal to 2, 1, and 0, respectively. Therefore, if the observed 
IBD score deviates from this mass function, it is an evidence that the observed IBD 
scores are not from a random model. Nevertheless, under specified condition the 
distribution of the IBD score is still worth investigating, especially to develop a more 
powerful test statistics or to do power analysis. 

Li and Sacks (1954) developed a method using stochastic matrices to derive the 
joint distribution of the genotype of two relatives. They used an example to demon- 
strate their method: a gene with two alleles, say A and a, with population frequencies 



18 



p and q = 1 — p. First, obtain a conditional probability matrix P={pij};, 




relative 2 
AA Aa aa 
Pii Pia P13 

P21 P22 P23 
P31 P32 P33 



where p n is the conditional probability conditional on relative 1 having AA, that 
relative 2 has AA; pu that relative 2 has Aa; and pi3 that relative 2 has aa, and so on. 
Once such a matrix is obtained, the absolute frequencies of all different combinations 
of genotypes of relatives pairs can be easily obtained by multiplying corresponding 
row by genotype frequencies of relative 1. In their example, that would be multiplying 
the first row, (pu,pu,Pi3) by p 2 , the second row by 2pq, and the third row by q 2 . Li 
and Sacks' method use three basic transition probabilities matrices, /, T, and 0, to 
construct P. The matrices, /, T, and is defined same as P except conditional on 
relatives pair having both, one, or no genes identical by descent, respectively. One 
can see that, in their example, 



/ = {»«} = 



1 











1 











1 



,r = {<y} = 



V 


o 





\p 


i 

2 








P 


<? 



,0 = {o l3 } = 



p 2 2pq q 2 
p 2 2pq q 2 
p 2 2pq q 2 



where / = 1,2,3 and j = 1,2,3. For the matrix /; since given that relative 1 and 2 
share two genes identical by descent, then the probability of the genotype of relative 
2 is AA, given the genotype of relative 1 is AA, is equal to 1, so i n = 1. For the same 
reason, the diagonal elements of /, in are equal to one, and off diagonal elements 
are equal to zero. Now consider the matrix T, since given that both relative share 
only one gene IBD, if relative 1 has genotype AA, and relative 2 also has genotype 
AA, relative 2 must receive one of his A alleles somewhere else other than the same 



19 

ancestor from whom relative 1 received his A allele. Since population frequency of A 
is p, thus tu of the matrix T is equal to p. The other elements of T and O can be 
obtained by the same reasoning. 

They showed several examples of how to use /, T, and O to construct P. The 
one for the parent-offspring relationship, the transition matrix P = T, and the 
grandparent-grandchild pair has transition matrix 

P = T 2 = -T + lo, 
2 2 

and for the general parent-offspring type relatives, 

P = T n+1 = l - n T + (l-( 1 -) n )0, 

where n + 1 is the number of generations between the two relatives. For the full-sib, 
the matrix is 

P = S=-I+-T + -0. 
4 2 4 

Campbell and Elston (1971) extended Li and Sacks' method to derive the transi- 
tion probability matrices for different modes of inheritance and multiple loci. Hase- 
man and Elston (1972) constructed the joint probability matrix for sibs. Risch (1990b) 
and Bishop and Williamson (1990) summarized these works and gave a table of trait 
IBD distribution conditional on marker IBD and relationship for sibs, grandparent- 
grandchild, uncle-nephew, half-sibs, and first cousins. It is adapted here in Table 2.3. 

Holmans (1993) showed that for an affected sib-pair, the possible values of prob- 
ability mass function p = (p ,Pi,l — Po — Pi) of marker IBD score 0, 1, and 2, is 
restricted in what they called "the possible-triangle", which is the intersection of 
Po > 0, pi < |, and pi > 2p , for all genetic models. Their proof is included. 

Let pj and z, be the probabilities of a sib-pair sharing ibd at marker locus and 
trait locus respectively, 6 be the recombination fraction between the marker and the 



20 



Table 2.3. The values of Pr(IBD of trait gene | IBD of marker and relationship). 



IBD of marker 


IBD of Trait 


2 


1 





sibs 


2 
1 



\]/2 

(l-*) 2 


2*(l-¥) 

(l-2* + 2* 2 ) 
2*(1 -9) 


(1-tf) 2 

$2 


grandparent-gra 


ndchild 






1 





1- 9 
9 


6 

1-9 


uncle-nephew 


1 





9(l-8) + \Q 

l- \e -9(i-$) 


l-\9-^{\-9) 
9(1-0) + \B 


half-sibs 


1 





i- * 


1-9 


first cousins 


1 





9(i-e) 2 + i$ 3 

§{i-|0 2 -*(i-0) 2 } | 


1- ±0 2 -*(l-0) 2 
{2+#(l-0) 2 + l0 2 } 



Where 9 is the recombination fraction between the marker and trait gene, 

* = 2 + (l -9) 2 . 



21 

trait gene, and tf = 9 2 + (1 - Of. Then, 

p = ^ 2 z + ^(l-^)z l + (l-^) 2 z 2 

Pi = 2*(1 -$)z + [y 2 + {l-y) 2 ]z 1 + 2tf(l -¥)z a 

p 2 = (l-f) 2 2o + *(l-*)2 1 + * 2 Z2. 

From the second equation, 

Pl = 2tf(l - *)(*„ + zi + z 2 ) + [tf 2 + (1 - tf) 2 + 2*(1 - *)]zj. 
Since j < $ < 1, and it can be shown that Z\ < i, 

Pl < 2^(1 - ^) + (2^ - l) 2 /2 = |. 
From the first and second equations 

Pi - 2p 
= (2* - 4tf 2 )z + (2* - \f Zl + {(1 - *)[2* - 2(1 - q)\}z 2 
= (2$ - l) 2 2l + 2(1 - *)(2* - 1)(1 - z x - zq) - 2#(2tf - 1)2 

> (2* - l) 2 ^ + 2(1 - *)(2* - 1) - 3(1 - tf)(2tf - l)2x - tf (2$ - 1)*, 
= 2(1 - *)(2* - 1)(1 - 2z x ) 

> 0. 

This proves the "possible triangle" restriction. He also proposed a procedure for find- 
ing the maximum likelihood estimator under the above restriction. That procedure 
is as follows: 

1. Obtain z = (zq, z\,1 — z — z\) from the unrestricted method which is sample 
frequency. 

2. If £i > |, remaximize z subject to the constraint z\ = |. If the resultant z > J, 
then reset z to null which is (0.25, 0.5, 0.25). 



22 

3. If 2z~ > z[, remaximize z subject to the constraint Z\ = 2z . If the resultant 
Zq > 45 then reset z to null. 

4. If z is in the triangle, leave it as it is. 

This concludes the procedure. 

Suarez, Rice, and Reich (1978) derived a generalized sib-pair IBD score distribu- 
tion in which the IBD score distribution is conditioned on the number of affected sibs 
for a trait with two alleles. This conditional distribution showed that "the distribu- 
tion of IBD scores depends only on the additive and dominance variances and the 
population prevalence of the disorder" (p. 94). 

One problem with using the IBD scores is, in order to establish IBD scores, marker 
alleles must be highly polymorphic. "Often, identity by descent cannot unequivocally 
be established" (Ott, 1991, p. 79). To overcome this problem with the IBD score, 
Lange (1986a) proposed an affected-sib-sets method based on IBS (identical by state) 
and at same time he extended the affected-sib-pair method to the affected-sib-set 
(ASS) method (Lange, 1986b). He proposed a test statistic for the affected sibs in a 
single nuclear family, 

Z = E** (2-1) 

1 t and j concordant (i.e. IBS=2), 

\ i and j half-concordant (i.e. IBS=1), (2.2) 

i and j discordant (i.e. IBS=0), 

where i and j are indexes of sib-set members. Lange noted that since they "will 
ultimately apply the Central Limit Theorem, it suffices to derive the mean and vari- 
ance of Z for a particular affected sib set." This ASS method uses population allele 
frequency to calculate the probabilities of all the possible parent mating types, then, 
conditional on the parental mating type, to calculate the probability of IBS of the 



Xu — 



23 

sib-pair. Lange gave the mean and variance of Z "in the limiting case of an infinite 
number of alleles, each of infinitesimal frequency," (this author believes that means 
very highly polymorphic) as follows, 

E(Z) = *(3-l)/4, (2.3) 

Var{Z) = a(*-l)/16, (2.4) 

where s is the size of the sib-set. He then presented a way to combine Z statistics 
from different sib-sets, 

T _ Y J sY,r W rs{Z rs -E(Z rs )) 

where Z rs denote the Z statistic for the rth affected sib set of size s, and w rs is 
a weighting. To assign a value to w rs , he claimed if the weights w rs depend only 
on s and not on r, and if the number of sib sets is large, then T should follow a 
standard normal distribution approximately. Based on Hodge's (1984) results, Lange 
recommended using the weights, 



(*-l) 



Var(Z rs )2 



In 1988, Weeks and Lange (1988) generalized the ASS method to pedigrees, calling 
the refined method the affected-pedigree-member (APM) method. In this method 
only those pedigree members who are both affected and typed at the marker locus 
enter the definition of test statistic. They modify the above test statistic T in two 
ways. First, modified the Z statistic to be 

Zij = -5{G ix , G JX ) + -6(G ix , G jy ) + -S(G iy , G jx ) + -S(G iy , G jy ), 

where i and j are the index for affected members in a pedigree, G, x and Gj x are the 
maternal marker allele for member i and j respectively, G !y and Gj y are the paternal 



24 



marker allele for member i and j respectively, and 

1 G and G' match in state 



8(G, G') 

G and G' do not match in state. 

They claim this modification "permits computation of the theoretical mean and vari- 
ance of the new test statistic for each pedigree by taking advantage of the theory of 
multiple-person ibd relation." Second, a new weighting factor 



V r m - 1 



y/Var(Z m ) 

where r m is the number of affected and typed individuals in the mth pedigree, and 
Z m is the Z statistic of the mth pedigree. Later, Week and Lange (1992) proposed a 
multilocus extension of this APM method. 

Bishop and Williamson (1990) studied the power of IBS methods on affected 
relative pairs and showed that several factors have a major influence on the power. 
These factors are, the relationship of the affected pairs, the polymorphism of the 
marker, the recombination distance between a trait locus and the closest marker, and 
the mode of inheritance of the trait. 

2.5 Risk Ratio 

In 1990, Risch published a series of papers addressing risk ratio methods. In the 
first of this series (1990a), a ratio of risk, Xr, is defined as the ratio of the probability of 
being affected given a type R relatives is affected, and the population prevalence. Let 
the relationship subscripts as follows: M=monozygotic twin; S=sibling (or dizygotic 
twin); 0= parent (or offspring); 2= second-degree relatives; 3=third-degree relative. 
For a single-locus model, two recurrence-risk pattern were derived; 



25 



A - 1 = 2(A 2 - 1) = 4(A 3 - 1) (2.6) 

and 

X M = 4A 5 - 2Ao - 1; (2.7) 

Similar results were obtained for multiplicative, additive, and genetic heterogeneity 
two-locus models and a general multilocus model. Important conclusions are: "for a 
single-locus model, Xr — 1 decreases by a factor of two with each degree of relationship. 
. . . For a multiplicative (epistasis) model, \ R - 1 decreases more rapidly than by a 
factor of two with each degree of relationship. Examination of \ R values for various 
classes of relatives can potentially suggest the presence of multiple loci and epistasis" . 
In the second paper (1990b), Risch derived the probabilities of IBD scores for 
a completely polymorphic marker for several relatives under the assumption that 
the recurrence-risk pattern (2.6) holds. According to his first paper of this series, 
this assumption might be violated if multiple contributing loci present. The IBD 
distributions were summarized in Table 2.4. He concluded that, "for diseases with 
large A values and for small 6 value, distant relatives offer greater power. For large 9 
values, grandparent-grandchild pairs are best; for small A values, sibs are best." The 
third paper (1990c) took into account the effect of marker polymorphism on power. 
There were some errors in the third paper and Risch wrote a response to address and 
corrected them (Risch 1992). The feasibility of his method is dependent on whether 
a suitable control group is included in the study or an appropriate estimate of general 
population risk can be obtained. 

2.6 Test Statistics Based on IBD Score 

There are several test statistics based on IBD score of the sib-pair method to detect 
linkage. One is the Two-allele (proportion) test statistic, which is the number 



26 



Table 2.4. The probability of IBD score for different relatives. 



Marker IBD 


Probability 


Affected sib pairs 


2 
1 



|+i£(2*-l)[(As-l)+2*(As-Ao)] 

|-i(2*-l) 2 ^(A 5 -Ao) 
i-£(2*-l)[(A 5 -l) + 2(l-*)(A5-Ao)] 


Grandparent-grandchild pairs 


1 



| + |(1 20) (Ao ' +1) (Ao 1) 
!-!(1-2*)7a£i7(Ao-1) 


Uncle(Aunt)-Nephew(Niece) 


1 



I + I(1-0)(1-20)^(A O -1) 
I-I(1-0)(1-20)^(A O -1) 


Half-sib Pairs 


1 



l + H2^-l)(^TT(Ao-l) 


First-cousin pairs 


1 



J + [(i - ey + e\i - ey + \e- - Jl^Ao - 1) 
| - [(i - ey + e\\ - ey + \e* - i]j^(Xo - 1) 



Where 6 is the recombination fraction between gene and marker, $ = 9 2 + (1 — 6)' 



27 

(or proportion) of sib-pairs that share two alleles, proposed by Day and Simons (1976) 
and Suarez, Rice, and Reich (1978). The second, the Mean test statistic, is the 
sum (or sample mean) of IBD scores, proposed by Green and Woodrow (1977). The 
third is to use the Pearson chi-square goodness-of-fit test (Pearson, 1900) on the 
distribution of IBD scores. All these tests assume that the IBD score frequencies are 
0.25, 0.5 and 0.25, under unlinked conditions, and using the deviation of observed 
IBD score from these frequencies to detect the linkage. 

Several papers discuss the merit of these test statistics. For the Two-alleles test 
statistics; the one proposed Day and Simons (1976) was based on the supposition that 
"the probability of sharing both haplotypes deviates more from the probability under 
the null hypothesis than does the probability of having one, or at least one, haplotype 
in common. One would test for the existence of the DS (disease susceptibility) gene 
by comparing the observed frequency of having both haplotypes in common." Suarez, 
Rice, and Reich (1978) suggested pooling categories IBD=0 and IBD=1, based on 
their generalized sib-pair IBD distribution that we mentioned in §2.4 "Distributions 
of IBD". They demonstrated that the affected-sib-pair that shared two alleles had 
the most information, noting that "with minimal loss of information, we can pool 
categories Pr{IBD=0,l}." 

Green and Woodrow (1977) used the total number of "repeats," i.e., the total 
number of IBD scores, as the test criterion. They also suggested using only affected 
sib-pairs because the distribution of IBD scores is symmetric under null hypothesis, 
so the sum of "repeats" can be approximated by normal distribution quicker than 
other sampling scheme. 

Blackwelder and Elston (1985) examined the Two-allele test statistic, the Mean 
test statistics, and Goodness-of-fit test statistic, by calculating the exact probabilities. 
They concluded that the Mean test statistic is generally more powerful than either 
of the other two. Schaid and Nick (1990) proposed a test statistic using a linear 



28 

combination of the number of marker alleles IBD as a test statistic. If the alternative 
IBD distribution can be specified, then this test statistic can be optimized. They also 
suggested a t max test statistic which was the maximum of the Mean test statistic and 
the Two-alleles test statistic and showed that t max was a more robust test statistic 
in the sense that, if the mode of inheritance (recessive, dominant, and so on) is 
misspecified, loss of power may not be too great. 

Faraway (1993) proposed a modified chi-square test. This test is a special case of 
a general method, x 2 test with restricted alternatives, given by Lehmann (1986, p. 
480). Faraway showed it was more powerful than the Two-alleles, the Mean, Pearson 
chi-square, or Schaid and Nick's t max tests for the finite sample and claimed it was 
the asymptotically uniformly most powerful invariant test. More details are given in 
the following: For testing hypothesis 

Ho : po - p 2 = 0.25, pi = 0.5 vs H x : p + pi + p 2 = 1. 

Let the chi-square goodness of fit test statistics be 

Y = An(p - 0.25) 2 + 2n(pl - 0.5) 2 + 4n(p 2 - 0.25) 2 , 

where pi is the sample frequency of IBD=i, n is the total sample size. Faraway showed 
that the possible value of IBD score distribution, />, i = 1,2,3, is restricted in region 
F that is the intersection of pi + j» 2 < 1, Pi < 1/2? an d 3pi/2 + p 2 > 1, and claimed 
this region is equivalent to the "possible-triangle" in Holmans (1993). Then, showed 
under this restriction, the {p^l ', p\l ', p2>) that maximizes statistic Y will actually turn 
Y into one of Two-allele test statistic, Mean test statistic, Y, and 0. The results are 
as follows. 



29 



Region 


Test statistic 


F 

2p 2 +P! >l,pi > 1/2 

3pi/2 + p 2 < l,p 2 > 1/4 

2p 2 +pi < l,p 2 < 1/4 


K 

Mean 

Two-allele 





All these discussions regarding the merit of different test statistics were based on 
a model where the trait locus has only two alleles. Since the distribution of IBD score 
was dependents on the penetrance of the genotype, as well as the frequency of the 
genotype, therefore, whether these results can be extend to a model that assumes 
three or more alleles for a disease susceptibility gene needs further investigation. 

Based on the same two-allele model, and Suarez, Rice, and Reich's (1978) gen- 
eralized sib-pair IBD score distribution, Knapp, Seuchter, and Baur (1994a) showed 
that if fl = /0/2, the Mean test was the uniformly most powerful test in 9 (recom- 
bination fraction) regardless of the allele frequency of t, where /o is the penetrance 
of the genotype tt (t is the susceptibility allele and T is the normal allele), /1 is for 
the genotype Tt, and ji is for the genotype TT. It is clear that recessive diseases 
(/1 = fi = 0) satisfied this condition, with either complete or incomplete penetrance 
(/o = 1 or /o < 1). The authors also proved the Mean test is the locally most powerful 
test (there is no test with a larger power for any alternative within a neighborhood 
of H according to Rao, 1973), irrespective of the mode of inheritance. He stated, 
"Because uniform optimality implies local optimality, no other test than the mean 
test can be uniformly most powerful." In another paper (1994b) Knapp, Seuchter, 
and Baur presented the equivalence of the mean test and the parametric maximum 
LOD score analysis with an assumed recessive mode of inheritance. 



30 

2.7 Likelihood Ratio Test 

The generalized likelihood ratio test has been commomly used in genetics for 
detecting linkage and heterogeneity. The LOD, an acronym for "logarithm of the 
odds ratio" Barnard (1949), is a special version of the likelihood ratio. However, the 
classic research in likelihood ratio test cannot be directly applied. Let's review this 
theorem first. 

In Serfling's book "Approximation Theorems of Mathematical Statistics", (1980, 
p. 144), he states the following: 

Regularity Conditions on F . Consider to be an open interval (not 

necessarily finite) in R. We assume: 

(Rl) For each 9 6 0, the derivatives . . . , 

where F is the family of distribution function, and is the parameter space of F. 
Assuming regularity conditions, it can be shown that the maximum likelihood es- 
timator of 9 has an asymptotic normal distribution (Serfling, 1980, p. 145). With 
this property and Lemma A (Serfling, p. 153), the likelihood ratio test theorem 
(Serfling, p. 155) can be proven as: 

Consider testing H : 9 = do, where 9 is a fc-dimensional vector in 0. Let 

L(9 ) 



A„ = 



^Peee 1 ^)' 



where n is sample size. Then, under H , the statistics -21n A n converge in distribution 
to xl (Serfling, 1980). 

The purpose of regularity conditions was to make sure of the existence of the 
Taylor expansion of the distribution functions (Serfling, p. 145). The parameter 
space was assumed to be an open interval in R, in fact, if 9 is an interior point of 
the parameter space 0, the theorem holds regardless of whether the parameter space 
is an open interval or not. 



31 



Let 

inn i«. max o<e<o.5 L{9) 

LOD = log 10 L{0 = o5) , (2.8) 

If the genetic model is known or is assumed to be known, some people use LOD, to 
to test 

H : 9= 0.5 vs Hf. 9 < 0.5, 

where 9 is the recombination fraction between a marker and a gene, and claimed 
21n(10)x LOD ~ Xi (Chotai, 1984; Elston, 1994; Shute, 1988). However, this is in- 
correct because 9q = 0.5 is on the boundary of the parameter space [0, 0.5]. Therefore, 
the maximum likelihood estimator does not have an asymptotic normal distribution. 
Similar errors happen when testing heterogeneity, where the parameter space is [0,1], 
and, under null hypothesis, monogeneity, i.e., a = 1 is on the boundary of parameter 
space (Ott, 1983), where a is the proportion of families belonging to the linked group. 
Ott (1991) attempted to correct this error, arguing that LOD is a "one-sided" test. 
Correct asymptotic distribution of maximum likelihood ratio under the condition that 
null parameter value is on the boundary of the parameter space was given by Self and 
Liang (1987). The asymptotic distribution of Eq. (2.8) is a 50:50 mixture distribution 
of a \\ an d a mass equal to 1 at 0. To make inferences, a critical value of LOD score 
needs to be decided. For simple Mendelian disease, the conventional criterion for 
claiming linkage is LOD > 3 (Morton, 1955, 1956). The probability of obtaining 
a LOD > 3 under the null hypothese is about 0.0001. One reason for such a low 
significance level is that the prior probability of that gene being within a certain 
distance from the marker is small; and another reason is that we would rather have 
no linkage map than have a wrong linkage map (Morton, 1955, 1956). Ott (1994) gave 
a Baysian argument, asking us to consider two hypotheses, H : free recombination 
or absence of linkage (9 = 0.5), and Hi: 9 < 0.5. The "prior" probability that a 
gene and a marker are within a measurable distance (40 CM) is small, and based on 



32 



Elston and Lange's result in 1975, he said the probability of a marker and a gene 
being within 40 cM was equal to 0.02. The posterior odds for linkage is, 

P(H x \data) ( P(data\ Hj) \ / P(#i) \ 
P{H \ data) ~ \P{data\ H ) ) \P(H )J ' 

Since, with LOD=3 the first term on the right-hand side is about 1,000:1 and the 
second term is about 1:50, the posterior odds against linkage are equal to 0.05. Since 
a multiple gene effect cannot be ruled out for a complex trait, he added one more 
hypothesis, the "other" hypothesis, H2. The prior probabilities are given as follows, 
with h being the prior probability of H2. 



Hypothesis Prior probability 

Ho: Single gene, no linkage 0.98(1-/*) 

Hi: Single gene, linkage 0.02(l-/i) 

H 2 : "Other" h 



The posterior odds are equal to, 



PjH^data) _ f P{data\H l ) \ I P(ff x ) 



P{H X \ data) " \P{data\ H Q ) J I pi H \ + \ P(data\H 2 ) l p(ff 

y v U/ P(data\ Ho) *■ 

where Th means "not Hi." Let R = fl^'l^i , then, 

1 P(data\ Ho) ' ' 

0.02(1 -h) 



) 



Q = 



0.98(1 - h) + hR ' 



If the disease in the family has an prior chance of 90% of being due to the "other" 
mode of inheritance, then the critical value has to increase from 3 to 4 in order to 
retain the posterior odds at 0.05. 

The above Ott's Baysian argument applies to a single marker situation. When we 
do a genome-wide search, we use many markers to do multiple tests. While the overall 



33 



false positive rate increases due to the nature of multiple tests, Ott (1991) indicated 
that, because the prior probability also increase with the number of markers, the 
critical limit could remain at 3. He also reported that Lander and Botstein (1989) 
investigated the problem and concluded that for the human genome, in order to keep 
the overall significance level at 5%, the appropriate critical level of LOD is between 2 
and 3. Later, Lander et al. (Lander and Schork, 1994; Lander and Kruglyak, 1995) 
strongly advocated higher LOD thresholds for genome-wide detection, results based 
on dense markers. This author have not tried to verify their "dense markers" approach 
hence will not include their results here, but would like to point out that for different 
methods the distribution of LOD are different. Thus, different thresholds should be 
applied to different methods as Lander and Schork, and Lander and Kruglyak did. 

In practice, it seems that most researchers forgot the other part of Morton's sug- 
gestion, that is, when LOD < -2, the marker should be excluded. This author has 
not seen a paper published using the method of exclusion. 

Clerget-Darpoux, Bonaiti-Pellie, and Hochez (1986) studied the effects of misspec- 
ified genetic parameters in LOD score analysis. They reported that "the power of the 
linkage test is sensitive to the degree of dominance, and slightly to the penetrance, 
but not to the gene frequency. In contrast, the estimation of the recombination frac- 
tion may be strongly affected by an error on any genetic parameter." MacLean et al. 
(1993) proposed a "MOD" method which is similar to LOD but it not only maximiz- 
ing LOD over recombination fraction but also maximizing over inheritance mode. 

2.8 Heterogeneity and Homogeneity 

There are two types of heterogeneity linkage analysis. One is allelic and the 
other is nonallelic; the latter is also referred to as locus heterogeneity. "With allelic 
heterogeneity, individuals differ from each other by having different alleles at the same 



34 

locus reponsible for the disease; in nonallelic heterogeneity, however, the disease is 
caused by different loci" (Ott, 1991). We will discuss only nonallelic heterogeneity 
in this section because there are more than one recombination fraction that can be 
detected by linkage analysis. 

There are several methods to test for homogeneity. These methods include a 
method proposed by Morton (1956), and a method by Smith (1963). Morton's method 
to test homogeneity is to test whether all the families have the same recombination 
fraction or have different recombination fractions. He proposed a simple statistic, 

k 

21n(10)x [£*(*)-*(*)]. (2.9) 

where 0, is the recombination fraction that maximize the LOD score of the ith family, 
and Zi(Oi) denotes the maximized LOD score, i = 1 to k. The Z(9) is the total LOD- 
score maximum that occurs at a value of for all families combined. This statistic is 
assumed to have a x 2 distribution with k - 1 df . 

Smith (1963) assumed that there were two groups, a linked group and an unlinked 
group. The linked group is the group of families that have a disease gene at a locus 
linked with the markers. The members of unlinked group either have no gene that 
caused the disease or have a gene at a locus that is unlinked to the markers. Smith 
proposed a test for testing whether all families in the study are from the linked 
group. If e is the true but unknown proportion of families belonging to the linked 
group with recombination fraction 0, let 6 and e be the MLE of e and 9. Under Hi, 
the LOD score of the ith family is given as Z,(e, 0) = log[eL,-(0) + (1 - £)£<(§)]. The 
total LOD score was equal to Z(e,6) = ^Z,(e,0). He assessed the significance of 
nonallelic heterogeneity by the statistic 

21n(10) x [Z(iJ\Hi) - Z(e = l,0\H o )]. (2.10) 



35 

If H is true, because e = 1 is on the boundary of parameter space, the statistic follows 
a 50:50 mixture distribution of a chi-square distribution with 1 df and a mass=l at 
0, not a chi-square distribution with 1 df as originally suggested. 

Ott (1983) compared Smith's method, which he called the A-test, and Morton's 
method, which he called the PS-test (M-test in his 1991 book). For a mixed situation 
(families with or without linkages between gene locus and marker locus), Ott com- 
pared the M-test with the A-test and found that the A-test was generally superior. 
However, since he erred on the distribution of the test statistics (2.10), his conclusion 
might need modification. 

Using IBD scores, Chakravart, Badner, and Li (1987) proposed a method to test 
linkage and homogeneity using IBD scores in the case of an autosomal recessive gene. 
The test was basically a goodness-of-fit test. They concluded that while the power of 
their method for detecting linkage from sib-pair data is excellent, that for detecting 
the heterogeneity of linkage is not. Proceedural details are as follows. 

For an autosomal recessive disease, the genotype of unaffected parents in multiplex 
families are Dd x Dd, where D and d are the normal and disease alleles, respectively. 
Among the affected sib-pair, let the probabilities of marker IBD =2, 1, and 0, be, 

P 2 = Pr(IBD = 2) = y 2 , (2.11) 

Px = Pr(IBD = 1) = 2xy, (2.12) 

Po = Pr(IBD = 0) = x\ (2.13) 

where x = 20(1 - 6) < 0.5 and y = 1 - x. The maximum likelihood estimator (MLE) 
for x is 

n\ + 2n 



x = 



2n 



(2.14) 



where n, are the numbers of affected sib-pair sharing i markers IBD, and n = £n,-. 
Because true value should not be greater than 0.5, they proposed a new estimator, 



36 



u, where, 



u 



x if x < 0.5 
0.5 if i > 0.5. 



This new estimator has a smaller variance under both null and alternative hypotheses 
than x's. The recombination value may be estimated by 



. 1 -y/l-2u 

= 



Under genetic heterogeneity, the IDB score distribution will be a mixture of two 
binomial distributions under 6 < 0.5 and 6 = 0.5 in the proportions c:\-c. Thus 
the /\'a in (2.11) through (2.13) become 

P 2 = c (l-x) 2 + (l-c)/A, (2.15) 

P x =2cx(l-x) + (l-c)/2, (2.16) 

P = cx 2 + (l-c)/4. (2.17) 

Solving the above equatons , they got c = p p jp P ^ P ■ Replacing P{ with its MLE, they 
obtained the MLE for c, and x under heterogeneity 

c = / n2 - no)2 , (2.18) 

n(n 2 — ni + no) 

_ n 2 - 2n 

Xh = Ti \ \ 1A ^> 

2(n 2 - n ) 

Then, they proposed a two stage method to test linkage and heterogeneity. First 
they tests linkage, if the null hypothesis of no linkage was rejected then they conclude 
heterogeneity. In the first stage, the statistic 

T = 2 v / (2n)(x - i) 



37 

was used and the authors claimed that T had asymptotic normal distribution. They 
did not explain why not using u. 
The statistic 

2 2 2 

G = n v + ^T^T — V + —2 - n 2 - 20 

n(l — xy lnx\\ — x) nx l 

is used to test heterogeneity. The authors claimed G is asymptotically distributed 
as a x 2 variable with df 1. However, they did not prove it nor point out what value 
should be used for x; use x or x~h. Monte-Carlo simulations were used to evaluate the 
power. 

2.9 Polygenes 

There are many traits in a population that have more variation and can not be 
categorized into distinct classes easily (Klug and Cummings, 1997). Traits exhibiting 
continuous variation may be controlled by two or more genes. Such traits are said to 
exhibit continuous or quantitative variation and are examples of polygenic inheri- 
tance. The hypothesis suggesting a large number of factors or genes were responsible 
for continuous phenotype are called the multiple-factor or multiple-gene hypoth- 
esis. These genes have a special name, polygenes (Griffiths et al., 1993). Klug and 
Cummings (1997, p. 95) summarized some major characteristics of multiple-factor 
hypothesis: 

1. Characters controlled by multiple- factors can usually be quantified by measur- 
ing, weighing, counting, etc. 

2. Two or more pairs of genes, located throughout the genome, account for the 
hereditary influence on the phenotype in an additive way. Because many genes 
may be involved, inheritance of this type is often called polygenic. 



38 



3. Each gene locus may be occupied by either an additive allele, which contributes 
a set amount to the phenotype, or by a nonadditive allele, which does not 
contribute quantitatively to the phenotype. 

4. The total effect of each additive alleles at each locus, while small, it approxi- 
mately equivalent to all other additive alleles at other gene sites. 

5. Together, the genes controlling a single character produce substantial pheno- 
typic variation. 

6. Analysis of polygenic traits requires the study of large numbers of progeny from 
a population of organisms. 

If there are more than two genes controlling the phenotype, then the number of 
genes involved and the effects of those genes are important to know. To address this, 
Tan and Chang (1972) proposed a method for estimating the number of genes for 
self-fertilized populations assuming that there were only two alleles for each gene and 
the effect of each gene is the same. This work was further expanded on by Tan and 
D'Angelo (1979) to estimate the numbers and effects of major genes and polygenes 
assuming that all major genes have the same effect and all polygenes have the same 
effect. Because their work were developed for self-fertilized population, we will not 
cover the details. 

2.10 Two-stage Genome Search 

Using the DNA-level genetic markers technology proposed by Botstein et al. 
(1980), Lander and Botstein (1986) concluded that the affected-sib-pair method can 
be used for a genome search for disease genes. A two-stage design, first a screening 
search to eliminate nonviable marker loci, and then an intensive search to identify 
gene location is an intuitive design. Although this design has already been used in 



39 

genome-wide searches, such as those by Davies et al. (1994) and Luo et al. (1995), 
no optimality of their designs was discussed. Thus, there were no guidlines on how to 
space the markers or how to allocate the available ASP in each stage. While Elston 
(1992, 1994) studied the "optimal" two-stage design and concluded that two-stage 
designs are more efficient than one-stage designs, he did not consider the statisti- 
cal complexity of the multiple test nor he considered the interval search nature in 
genome-wide linkage detection or the resource constraints. In another paper, Brown 
et al. (1994) studied the multiple-stage approach for genome search using affected- 
pedigree-member, by simulation. Since Brown et al. used only simulation and only 
one pedigree was considered, it is not obvious how to apply their results to the 
ASP method. Darvasi and Soller (1994) have studied the optimal spacing of genetic 
markers for the QTL trait without considering the two-stage approach. Holmans 
and Craddock (1997) conducted a simulation studied on the efficient strategies for 
genome scanning using maximum-likelihood affected-sib-pair analysis. The situation 
they considered are: a 200 affected sib-pairs sample, four different sample allocation 
strategies, and five grid-tightening strategies. The risk ratio of sibs are equal to 2 or 3, 
each marker locus have five equi-frequent alleles, and there are five possible location 
of genes. Since their studies were simulation studies, it is not clear how to generalize 
their results. Furtheremore, they did not consider different sample size, nor optimal 
strategies under resource constrains. 

The main goals of this dissertation is to answer the two stage design question as 
they apply to rare autosomal recessive and dominant diseases with a dichotomous 
phenotype (affected or not affected). 



CHAPTER 3 
TWO-STAGE GENOME SEARCH FOR SIMPLE MENDELIAN DISEASE 



In a two-stage search method, the first stage uses part of the ASPs with a wide 
spread markers in the genome. The rest of the resource is to be used on those 
promising markers identified in the first stage. In this study, only the least favorable 
configuration, that gene that lies in the middle of two adjacent markers, will be used 
for constructing the designs. 

There are three major statistical problems involved in deriving analytic solutions 
are summarized as follows: 

1. Because all the markers on the same chromosome are linked, the IBD scores 
are not independent. In the case of a genome search, there are many markers 
spread along the chromosomes. We have to handle a high-dimensional joint 
distribution. 

2. In the first stage, a number of loci will be chosen for the more intensive linkage 
analysis in the second stage. The number and position of these loci that pass 
to the second stage are random. This makes calculation of exact distribution 
in the second stage extremely difficult. 

3. Since in the first stage only a small number of ASPs and a large number of 
markers are used, there will be many ties in the IBD scores. The asymptotical 
solutions using a continuous normal distribution cannot handle these ties. 



40 



41 



In this dissertation, Problem 3 was handled by a multinormial distribution. We used 
independent model as an approximation for Problem 1 and 2, and checked the results 
with simulation. 

3.1 Assumptions 

1. There is one and only one disease gene that increases the probability of an 
individual being affected. However, the disease may have nongenetic causes. 

2. Highly polymorphic equally spaced markers are available. When m markers are 
assigned in the first stage, their positions are £- •&-, .... ( 2t " -1 ) l where L, the 

° ° ' r 2m 2m ' 2m ' 

length of the genome, is equal to 3300 cM (Lewin, 1990). 

3. For a rare autosomal recessive disease, the parents' disease genotypes are D\d 
and D 2 d; and for a rare dominant disease, the parents' disease genotypes are 
Did and D 2 D 3 , where d is the disease gene. 

4. In the same family, the probability that the disease of one affected sib is caused 
by a gene and the other is not is negligible. 

5. The cost of typing alleles is a constant, i.e., the cost of typing A; markers from 
one person is the same as typing one marker from k persons. This assumption 
of cost ratio can be relaxed to suit practical situations, but the numerical result 
would need to be re-calculated. 

6. We use the Two-alleles statistic (Day and Simons, 1976), i.e., let X i} =l if ith 
marker IBD score = 2 for the jth sib-pair, and Xij=0 otherwise. In our analytic 
approach, the Xij are assumed to be independent random variables, except the 
two X{j adjacent to the gene. 



42 

3.2 Two-Stage Procedure 

Suppose there are n ASPs and there are enough resources to type N marker loci. 
Three numbers need to be determined in a two-stage design: n\ and m, the number 
of ASPs and the number of markers to be used in the first stage (Stage I), and r, the 
number of markers to be used in the second stage (Stage II). The markers chosen for 
the second stage are based on the statistic used by Day and Simons, 

wi = 2_^ X*ji (3-1) 

in the first stage, where X{j is defined in Assumption 6. Ideally, the r markers with 
the highest scores are chosen for Stage II. However, in the event of ties, more than r 
of them may have to be chosen. Thus, #, the actual number of markers used in Stage 
II, is a random variable. The formal definition of R is: R > r, but if the marker(s) 
with the lowest score in this group (markers for stage II study) is (are) taken away, 
then the total number of remaining markers is smaller than r. 

In Stage II, R markers on N 2 ASPs are to be typed, where N 2 is the largest 
number of sibs that can be used subject to the resource constraint. Since R is a 
random variable, N 2 is also a random variable. Without loss of generality, let the R 
markers be 0,1, ..., R—l, and the sib-pairs in the Stage II be ni + l,ni+2, ...,ni + N 2 . 
Thus, N 2 is the largest x such that rani + Rx < N. Define 

ni+N 2 

2 Si= Y, X *i (3-2) 

j=ni+l 

Then the marker with the uniquely highest 2 S{ is claimed to be the locus nearest the 
disease gene. If two adjacent markers have the same highest score, then the gene is 
claimed to lie between them; otherwise, the gene location is undetermined. In this 
case, the two markers are called a marker group. 



43 

Once the location, /, has been chosen, say I = i, the next step is to check whether 
we can claim linkage. Let < Q ,„ 2 ,/? be the 100(1 - a) percentile of the unique maximum 
of R binomial (n 2 , 0.25) random variables. If 2 S/ > £ Q ,n 2 ,.R> then we claim there is a 
linkage at loci i at significance level a. 

3.3 Probability of Allocating the Correct Marker 

3.3.1 Analytic Approach 

To make the analytic solution tractable, Assumption (6) is made on the distribu- 
tions of Xij. They are later compared with a more realistic situation by simulation. 

Let markers be numbered from to m — 1 . Without loss of generality, if the gene is 
at the end of the chromosome, we let the nearest marker be marker 0; and if the gene 
is between two markers, we let it be located in the middle of marker and 1. If we 
assume that gene location is uniformly distributed along the genome, the probability 
of the gene at the end is 1/m. 

Let iS( r ) be the rth largest statistic in the set {iS,-} in the first stage. The 
probability of finding the gene by the two-stage method is, 

F(The marker closest to the gene is found) 
= F(Gene is at the end of the chromosome, marker is chosen in Stage II) (3.3) 
+ P(Gene is not at the end, either marker or 1 is chosen in Stage II). (3.4) 

Eq. (3.3) can be written as a sum of the probabilities of three exclusive events; 
Eq. (3.5) through Eq. (3.7). 

F(Marker is chosen at the end of Stage II | gene is at the end) 



m 

1 
m 



P(i5 passes Stage I and 2 So is the highest in Stage II) 



44 



1 ni 
= — > PiiSo passes Stage I, 260 ' s the highest in Stage II, iS( r _D = k) 
m ^— ' v ' 



fc=o 



m 



1 ■» - 
— 2J {■Pd'S'o > iS( r - 1)1 i5( r _j) = k, 2 S is the highest in Stage II) (3.5) 



fc=o 



+ P[k = i5( r _j) > \Sq > i5( r ), 2<5o is the highest in Stage II) (3.6) 

+ P{k = i5( r _!) > i5 = i5( r ), 2^0 is the highest in Stage II)} . (3.7) 



Eq. (3.5) is equal to, 



P(iSo > i5( r _!), l&r- 1) = ^)2'S'o is the highest in Stage II) 
(assuming independence) 

E £ 

i i P{iSo > k, i iS/s > k, I \Sj's = k, 

, + l > (r-l), i < (r-1) 

(m - 1 - i - I) iS/s < k, 2 So > max 2 S t ) 

(#0 



r — 2 m — 1— i 



r — £. in—i — i f 

" E E P USo>A;) 

i=0 fcr-l-! *■ 



m 



1! 



t!/!(m-l -»-/)! 



v/ 



P(i5j > k) 1 P^Sj = k)' P^Sj < k) 

N 2 



m — l—i—l 



t=l 



Y,P(2S = t)P( 2 s J <ty + < 

N — n\m 



where N 2 



i + / + 1 



Int 



Eq. (3.6) is equal to, 



P{iS( r _x) = k > iSo > iS( r ), gene is found) 



Jfc-2 



= /]P{lS( r -i) = k > x5 > i5( r j = z, 2 So is uniquely highest) 



t=0 



fe-2 r-l (m-l)-(r-l) 

= EE E P (* = i 5 (-i) > 1^1 > i5(r) = *, (r - 1 - lS/« > *> 

i=0 1=1 /i=l 



45 



I iS/s s jfc, ft iS/s = •*, (m - r - ft) iS/s < t, 2 S'o is uniquely highest) 

£2 «-l HH'-i) j ^^ 

~ E E 2_. \ P(A;>l5o>z) ( r -l -/)!/! ft! ((m-l)-(r- 1)- ft)! 

1=0 /=1 /i=l 

PiiSj > ky- 1 - 1 PitSj = *)' P(tSj = i) h p{iS 3 < ir -i-(r-D-i 
Y^p( 2 s = t)P( 2 s J <ty- 1 



t=i 



where N 2 = 



J Int 



Eq. (3.7) is equal to, 

F(i5 (r _i) = k > iSq = i5( r ), gene is found) 



k-i 



= ^P(iS( r _!) = A; > 1S0 = i5( r ) = i, 2^0 is uniquely highest) 

1 = 

k-l r-1 (m-l)-(r-l) 

= S J2 H p ^ k = iSr ~ i > iSi = i5r = ^ ( r - x - 1 5 / S > fc > 

,=0 (si /i=l 

/ jSy'a = fc, h iS/s ss t, (m- r - h) iS/s < i, 2 S is uniquely highest) 

m- 1! 



fc-l r-l (m-l)-(r-l) 



(r - 1 - /)! /! ft! ((m - 1) - (r - 1) - ft)! 
t=0 /=1 h=l v f \\ t \ t 1 

PiiSi > ky- 1 -' p(iSj = k) 1 PdSj = i) h p(iSj < £)«*-i-(r-i)-i 

■N a 



t=\ 



Y,P(2S = t)P( 2 s J <ty- 1+h 

N — n\m 



where N% = 



Eq. (3.4) can be written as 



r + h 



Int 



771—1 
771 

771—1 
771 



F(Marker or 1 is chosen at the end of Stage II | gene is not at the end) 
{P(iSq passes Stage I and 1S1 does not, 2S0 i s tne highest in Stage II) (3.8) 



46 



+P(i5i passes Stage I and i5o does not, 2-51 is the highest in Stage II) (3-9) 

+P(\So and i5i pass Stage I and 2S0 or 2S1 is the highest in Stage II)}. (3.10) 

Since the gene is assumed to be in the middle of two markers, Eq. (3.8) is equal 
to Eq. (3.9) and they 

= P(\So passes Stage I and i5i does not, 2^0 is the highest in Stage II) 
= \_]P(iSq passes Stage I and i5i does not, 2S0 is the highest in Stage II, 



k=\ 



Til 



iS( r _i) = fc) 



= Yj {P{iSo >i5( r _!) = k > jSi, 2-50 is the highest in Stage II) 



(3.11) 



lbs] 



+ F(i5( r _ 1 ) = k > \Sq > i5( r j, i5o > 1S1, 2S0 is the highest in Stage II) (3.12) 
+ P(i5( r _!)=A; > iSq = i5( r ), i5o > iS\, 2S0 is the highest in Stage II) }(.3.13) 



Eq. (3.11): 



P{iSo >i<S(r-i) = k > i5i, 2S0 is the highest in Stage II) 
P(i5 > ac, i tS'jS >k ,1 ify = fc, 



EE 

i / 

i + l > (r- 1), . < (r- 1) 



(m-2-i-l) iS';S < k, x 5i < k 2 5 > max 2 5 t ) 



r-2 m-2-t , 

E E 

t'=0 ;=r-i-» v 



P(iS >*,iSi<*)- 



m-2! 



»!/!(ro-2-*-Z)! 
P(i5j > fc) 1 PfcSj = fc)' P(i5> < fc)" 1 " 2 -'-' 

2 

^P( 2 5o = i)P( 2 5, <t)«'+' 

iV - nim~ 



N. 



L t=l 



where N2 



i + l + l 



Int 



47 



Eq. (3.12) is equal to, 

P(k =i S( r _!) > i5 > iS (r ) = t, iSq > iSi, 2S0 is the highest in Stage II) 

k-2 r-l (m-2)-(r-l) 

= *P T^ ^ P(& = iS( r _i) > 1S0 > i5( r) = i, i5 > 1S1, 

t'=0 (=1 /i=l 

(r - 1 - I) iS/s > k, I iSj's = k, h iSj» = », (m - r - 1 - h) t 5/« < i, 

2 5 > max 2 S t ) 
</o,i 

fc_2 r-l (m-2)-(r-l) 

= E E E 



=0 /=1 /l=l 



m-2! 



|p(* > x5 > i, 1S0 > i5i) (r _ 1 _ /)!/! w((m _ 2) _ (r _ 1) _^ )! 
P(i5i > ^) r - x -' P(iSj = A)' Pd5, = 0* P(iSj < i) m - r ~ l - h 

N 2 

Y / p( 2 s = t)P( 2 s J <ty- 1 

N — n\m 



t=\ 



where iV 2 — 



Int 



Eq. (3.13) is equal to, 



k-2 



^P(fc =1 5 (r _ 1) > iStj = 1%) = », i5 > i5i, 2 5 is the highest in Stage II) 



i=i 



fc-2 r-l (m-2)-(r-l) 

^ ^ 2J f( fc = lS(r-l) > 1^0 = lS(r) = »I 1^0 > l5l, 

i=l /=1 /i=l 

(r - 1 - /) 1 S/s > fe, / i5/s = k,h iS/s = i, (m - r - 1 - h) iSj's < i, 

2 5 > max 2 5i) 
^0,1 

k-2 r-l (m-2)-(r-l) 

E E E 

t=l 1=1 h=l 

lp(iS = i, l Si<i) 



(r-l-/)!/! /i! ((m-2)-(r-l)-/i)! 



48 



PiiSj > k) r ~ 1 - 1 P{ t Sj = k) 1 P^Sj = i) h PixSj < i) 



m—> — 1 — h 



Ni 



t=\ 



J2P(2So^t)P( 2 s J <ty- 1+h 

N — n\m 



where N 2 



r + h 



Int 



Eq. (3.10) is equivalent to 

F(i5o and iSi pass Stage I, gene is found in Stage II) 
= >J{F(i5o >i5( r _ 2 ) = k, iS\ >i5( r _2) = k, gene is found in Stage II) 



+ P( 


+ P( 


+ P( 


+ P( 


+ P( 


+ P( 


+ P( 


+ P( 


+ P{ 


+ P( 


+ P( 


+ P( 



iS >i5( r _ 2 ) = k > jSi > i5( r _x 
i5o >i5( r _ 2 ) = k > i5i = i5( r _! 
i5i >i5( r _ 2 ) = k > iS > i5( r _x 

l5i >l5( r _ 2 ) = k > jSo = l^r-i 

i5( r _ 2 ) = k > x5 > i5i > i5( r _ 
i5( r _ 2 ) = k > iSo > i5i = i5( r _ 
i5( r _ 2 ) = k > i5i > i5o > i5( r _ 
i5( r _ 2 ) = k > iS\ > iSq = i5( r _ 
i5( r _ 2 ) = k > i5i = i5q > i5'( r _ 



, gene is found in Stage II) 
, gene is found in Stage II) 
, gene is found in Stage II) 
, gene is found in Stage II) 
), gene is found in Stage II) 
), gene is found in Stage II) 
j, gene is found in Stage II) 
), gene is found in Stage II) 
w gene is found in Stage II) 
), gene is found in Stage II) 



i5( r _ 2 ) — k > iS\ — \Sq — i5( r _ 

i5( r _!) = k > \S\ — iSq > iS( r ), gene is found in Stage II) 

iS( r _!) = k > iS\ = iSq = i5( r ), gene is found in Stage II)} 



3.14) 

3.15) 
3.16) 
3.17) 
3.18) 
3.19) 
3.20) 
3.21) 
3.22) 
3.23) 
3.24) 
3.25) 
3.26) 



Again, because the gene is in the middle of two markers, Eq. (3.15) is equal to 
Eq. (3.17), (3.16) is equal to (3.18), (3.19) is equal to (3.21), and (3.20) is equal to 
(3.22). 



49 



Eq. (3.14) is equal to, 

P(iSo > i5'( r _2) = k, 1S1 > i5( r _ 2 ) = k, 2 S is uniquely highest in Stage II) 
+ P{iSo > i5( r _ 2 ) = k, 1S1 > i5( r _ 2 ) as k, 2 S\ is uniquely highest in Stage II) 
+ P(iS > i^(r-2) = k, 1S1 > i5 (r _ 2 ) as fe, 2 5 ss 2 Siis the highest) 
= P{iSo > i5'( r _2) = k, 1S1 > i5'( r _2) «= fc| 250 or 2 Si is uniquely highest in Stage II) 

* v ' 

(A) 

+ P{iS > i5 (r _ 2 ) = k, iSi > i5 (r _ 2 ) = k, 2 S = 2 S{is the highest) 



(A) 



(B) 



E E 

• / P(iS > k, i5i > A;, i5( r _ 2 ) = fc,j iS/s > k, 

i + l > (r-2), • < (r-2) 

/ iS/s = fc, (m — 2 — z — /) iS/s < Ac, 2-50 or 2S1 is uniquely highest in Stage II) 

r— 3 m—2—i 



r—o m — L — i f 

= Yl £ {j*(i&£*iiSi£*); 

t=0 /=r-2-i *■ 



m-2! 



i!/! ((m-2) -»'-/)! 
P(i5> > fc)' P(i5j = fc)' P(i5j < k) m - 2 - z ~ l 



P( 2 S > max 2 5(,25'o > 2 Si) + P{ 2 Si > max 2 S<, 2 Si > 2S0) > 
</o,i (#0,1 J J 



r — 3 m — 2 — i 



r — o m — 4 — 1 • 
1=0 /=r-2-! ^ 



m-2! 



i\l\ {{m-2)-i-l)\ 



p{ x Sj > ky p^Sj = ky PdSj < ky 

N 2 



-2-i-l 



J2 p (2S = t, 2 s 1 <t)P( 2 s J <ty+ l 

N — n\m 



t=\ 



1 



where N 2 = 



lr + l + 2\ In t 



(B) 



EE 



i / P(iS > k, 1S1 > k, iS( r _ 2 ) — k,i iS/s > k, 

i + l > (r-2), 1 < (r-2) 



50 



/ yS/s = k, {m-2-i-l) iS/s < k, 2 S = 2 Si is uniquely highest in Stage II) 



r-3 m-2-i 



r-o m-t-i , m-2' 

= £ £ p^.A^) ,, f!((m _ 2) I,.,), 



i=0 l-r-2- 



p^Sj > ky pus, = k)< PdSj < ky 



P^So = 2S1 > max 2 S t 

t#0,l 



r— 3 771— 2— t /• 

£ £ |p(i5 >fc,i5i>A) 

:—f\ 1 n ." l> 



i-2-i-l 



m-2! 



i!/! ((m-2)-t-/)! 
P(i5j > *)' P(iS> = *)' P(iSj < k) m ~ 2 -'- 1 

J2 p (2So = t, 2 S l =t)P( 2 S J <t) r+l 

N — n\m 



t'=0 l=r-2-i 



t=\ 



where N 2 = 



[r + l + 2\ Int 



(A) + (B) 



r — 3 m — 2—i s 

E E 

i=0 l=r-2-i K 



P{iS >k,iSi>k)- 



m-2! 



t!/! ((m-2) -i-/)! 



PdSj > lb)' P(i5j = ky PdSj < ky 

N 2 



J][2P( 2 5o - t, 2 5i < + P( 2 5 = t, jS, = t)]P( 2 5, < 

JV — nim 



r+J 



. fc=l 



where iV 2 



L r + / + 2 J /Tl( 



Eq. (3.15) and (3.17) are equal to 

P{iS > i5 (r _ 2) = k > 1S1 > i5( r _!), gene is found) 

fc-2 
= ^{P(i5 > iS( P _ 2 ) = fc > i5i > i5( r _ x) = i, 2 S or 2 S X 



1=0 



is uniquely highest in Stage II) 



51 



+P(\So > i5( r _ 2 ) = k > i5i > i5( r «i) a t, 2'S'o = 2-51 is the highest in Stage II) 

k-2 r-l (m-2)-(r-2) 

= 53 X] E {^(i^o>*,ib>i5i>t,(r-l-0i5/«>Jb J 

/ x5j's = k,h iS/s = i, (m — r — 1 — h) \S 3 ' s < i, 2S0 or 2S1 
is uniquely highest in Stage II) 
+F(i5 > k, k > i5i > i, (r - 1 - /) L 5/« > k, 

I iS/s = k,h iS/s = i, (m — r - 1 - h) iS/s < i, 2S0 = 2S1 
is the highest in Stage II)} 

k-2 r-l (m-2)-(r-2) 

= EE E 

i=0 /=1 h=l 



P{ 1 S >k,k> 1 S 1 >i) 



m-2! 



(r - 2 - /)! l\ h\ ((m - 2) - (r - 2) - h)! 
P(i5 ; > fc) r - 2 -' PdSj = k)' P{iSj - i) h PixSj < i)»-J-MH 



JV 2 



J](2P( 2 5o = t, 2 S 1 <t) + F( 2 5 = t, 2 5i = t))f feSj < t)*" 

iV — Tiim 



i=l 



where N2 = 



Int 



Eq. (3.16) and (3.18) are equal to 

P(iSo > i5( r _ 2 ) = k > 1S1 = iS( r _i), gene is found) 



fe-l 






= 2_^{P{iSo > i5'( r _2) = * > 1S1 = i5( r - 1) = i, 2«S"o or 2S1 



1=0 



is uniquely highest in Stage II) 
+ P{iSq > i5( r _ 2 ) = k > x 5i = i5( r _!) = i, 2 5o = 2S1 is the highest) 

k-l r -2 (m-2)-(r-2) 

= J2Y1 Yl {P(iS >k, 1 S 1 >i,(r-2-l) l S J 's>k, 

i=0 1=1 h=l 

I iS/s — k,h iS/s — i, (m - r - h) \Sj's < i, 2 So or 2 S'i 



52 



is uniquely highest in Stage II) 
+ PdS > k, i5i >= t, (r - 2 - /) i5/s > fc, 

/ i5/* = *, h iSj'a = i, {m-r- h) iS/s < i, 2 S = 2 Si is the highest)} 

k -l r -2 (m-2)-(r-2) 

EE E 

i=0 /=1 /i=l 



P(i5 > fc, i5i > 



m 



2! 



(r-2 - /)! /! M ((m - 2) - (r - 2) - /»)! 

P(l5j > k) r - 2 - 1 PdSj = k) 1 PdSj = i) h P(lS 3 < i)m-2-(r-2)-l 



N 2 



J](2P( 2 5o = t, 2 5i < + P( 2 S = t, 2 S 1 = t))P( 2 Sj < t) 

N — n\m 



r-2+h 



. (=1 



where iV 2 = 



r + h 



Int 



Eq. (3.19) and (3.21) are equal to 

P(i5 (r _ 2) =k>iSo>iSi> i5( r _i), gene is found) 



fe-l 



= ^{P(i5 (r _ 2 ) = k > iSo > 1S1 > i5( r _!) = i, 2 5 or 2 5 x 

t'=0 

is uniquely highest in Stage II) 
+ P(i5( r _ 2 ) = k > iS > i5 x > i5( r _ 1} = *, 2 5 = 2 5iis the highest)} 

A:_ 3 r-2 (m-2)-(r-2) 

= EE E 

;=0 ;=i h=i 

{P{k > t Sf> > i5i > i, (r - 2 - /) t S/» > k, I iS/s = k, h iS/s a i, 
(m - r - /i) iS/s < t, 2 5o or 2 Si is uniquely highest in Stage II) 
+ P{k > i5 > i5i >= *', (r-2- /) iS/s > k, 

I tSj's = k,h iSj's = i, (m- r - h) iS/s < i, 2 S = 2 Siis the highest)} 

fc-3 r-2 (m-2)-(r-2) 

= EE E 

t=o ;=i h=i 



53 



P(k > iSo > iSi > i) 



m-2! 



(r - 2 - /)! /! ft! ((m - 2) - (r - 2) - ft)! 
P(iSj > k) r ~ 2 - 1 PdSj = k) 1 P{xSj = i) h P(iSj < i)— 2-(r-2)-/ 1 
N 2 

J](2P( 2 5o = t, jS, < + P( 2 S = t, 2 5j = t))P( 2 S 3 < t) r ~ 2 

N — n\m 



t=i 



where iV 2 



Eq. (3.20) and (3.22) are equal to 



r J Int 



P(i5( r _ 2 ) = A; > i5 > i5i = i5( P _i), gene is found) 



Jfc-2 



2Z{-f > (i'5(r-2) = A; > i5 > i5i = i5( r _!) = i,aSo or 2 Si 



i=0 

is uniquely highest in Stage II) 
+ P{iS( r _ 2 ) = k > \So > iSi — i5( r _!) = r, 2-S'o = 2S1 is the highest in Stage II)} 

k-2 r-2 (m-2)-(r-2) 

= EE £ 

t =o 1=1 h=l 

{P(k > 1S0 > i, 1S1 = i, (r - 2 - /) t 5/« > fc, / i5/« = k, h iS/s = i, 

(m — r - h) \Sj's < t, 2 So or 2 S\ is uniquely highest in Stage II) 
+ P(* > 1S0 > •', 1S1 = i, (r - 2 - /) iS/a > Ik, / i$/a = fc, /1 iS/s = i, 
{m — r — h) iS/s < i, 2 5o = 2 5iis the highest in Stage II)} 

k-2. r-2 (m-2)-(r-2) 

- EE £ 

t=0 Izzl h=l 



{P{k> iS >i,iSi = i) 



m 



2! 



(r-2-l)\ll ft! ((m - 2) - (r - 2) - ft)! 

PiiSj > k) r ~ 2 -' P^Sj = k) 1 P^Sj = l) h P{ X Sj < i)m-^-{r-2)~h 
N* 

J](2P( 2 5o = t, 2 S 1 <t) + P( 2 S = t, a Si = 0) x P( 2 S J < t) r ~ 2 + h 



54 



where N 2 = 



N - n\m 
r + h 



Int 



Eq. (3.23) is equal to 

P{iS( r -2) = k> r So = 1S1 > i5 (r _!), gene is found) 

k-2 

= ^{Pd5 (r _ 2) = k > i5 = 1S1 > i5( P _i) = *, 2S0 or 2 Si 



t=0 



is uniquely highest in Stage II) 
+ P{\S {r -2) = k > 1S0 = 1S1 > i5( r _!) = i, 2 5 = 2 5i is the highest in Stage II) 

k -2 r-2 (m-2)-(r-2) 

EE E 

{P(/fc > 1S0 = 1S1 > *,(r - 2 - /) iS/s > k, I iSj's = fc, ft i5/s = i, 
(m - r - h) iS/s < t, 2 5 or 2 Siis uniquely highest in Stage II) 

+ P(k > iSo = i5\ > i, (r - 2 - /) iSj's > k, I x Sj'a = k,h iSj's = i, 
(m-r - h) iS/a < i, 2 S = 25iis the highest in Stage II)} 

k-2 r-2 (m-2)-(r-2) 

EE E 

t =o ;=i /i=i 



P(fc > 1S0 = i^i > 



m-2! 



(r-2 - /)! I! h\ ((m - 2) - (r - 2) - /*)! 

P^Sj > ky- 2 ~ l P^S, = k)' PdSj = i) h P{lSj < i)m-2-(r-2)-h 

N 2 



£(2P( 2 S m t, 2 Si <t)+ P( 2 S = t, 2 5i = OJWj < t) 

N — uitti 



r-2 



t=l 






where iV 2 = 



r J/nt 



Eq. (3.24) is equal to 



P(i5 (r _ 2) = * > 1S0 = 1S1 = i$(r-i), gene is found) 



55 



k-i 



= ^{F(i5 (r _ 2) = k > iSo = 1S1 = i5 (r _ 1} = i, 2 S Q or 2 Si 



i=0 



is uniquely highest in Stage II) 
+ P(i5 (r _ 2) = k > iSo = 1S1 = iS( r -i) = i, 2S0 = 2S1 is the highest in Stage II) 

fc_l r _ 2 (m-2)-(r-2) 

= J2 J2 £ {P(i5 = iS 1 = t,(r-2-Oi5 j 's>*r, liSj's = k, 

i-0 1=1 fcsl 

/i i5/s = *, (m-r- /i) i5/s < t, 2 5 or 2 Si is uniquely highest in Stage II) 
+ P(i5 = i5i = t, (r - 2 - /) iS/i > k, I jS/s = k, 

h iSj's = i, (m-r-h) iS/s < i, 2 5 = 2 Si is the highest in Stage II)} 

fe-l r-2 (m-2)-(r-2) 

= EE E 



;=o ;=i /i=i 



m-2! 



|p(i5 - i5i - (r _ 2 _ qj „ w ((m _ 2 ) _ ( r _ 2 ) - ft)! 

PdSj > k) r - 2 -' PbSj = *)' P(l5, = i) h PdSj < i)m-2-(r-2)-h 



N 2 



J](2P( 2 5 = t, 2 5i < + P( 2 5 = t, 2 Si = t)) x P( 2 5, < t) 

N — n\m 



r-2+h 



1=1 



where iV 2 



r + h 



J Int 



Eq. (3.25) is equal to 



P(iS( r _ X ) = A; > i5 = i5i > i5 (r) , gene is found) 

fc-2 

^{P(iS (r _i) - fc > 1S0 = i5i > i5 W = i, 2 5 or 2 5i 






i=o 



is uniquely highest in Stage II) 
+ P(i5( r _!) = fc > i5 = i5i > i5 (r) = i, 2 S = 2 S\ is the highest in Stage II) 

fc-2 r-1 (m-2)-(r- 1) 

= EE E 

i=0 (=1 h=l 



56 



{P{k > iSo = 1S1 > i, (r - 1 - I) iS/s > Jfc, / iS/s = k, h ,5/« = »', 
(m — r — h) iS/s < i, 2 So or 2 5i is uniquely highest in Stage II) 
+ P(k > 1S0 = i5i > i, (r - 1 - /) ^/s > k, I x 5/s = k, 

h iS/s = i, (m - r — h) iS/s < i, i5i = 2-S'iis the highest in Stage II)} 

fc — 3 r-l (m-a)-(r-l) 

= EE E 

1=0 /=i h=i 



P{k > iS = i5i > i) 



m-2! 



(r-l - /)! l\ h\ ((m - 2) - (r - 1) - /*)! 
P{iSj > ky- 1 -' PdSj = k) 1 P(tSj = i) h PdSj < j)— 2-(r-D-^ 

5^(2P( 2 5 = t, 2 S 1 <t) + P( 2 S = t, 2 S 1 = t)) x P{ 2 Sj < t) r ~ l 

N — n\m 



t=i 



where N 2 



r+ 1 



Int 



Eq. (3.26) is equal to 

P(i5 (r _ 1 ) = k > 1S0 = 1S1 = iS( r ), gene is found) 

fc-i 
= 2_^{P(iS( r -i) = k > 1S0 = 1S1 = i5( r ) = i, 2 S or 2 5 x 

i=0 

is uniquely highest in Stage II) 
+ F(i5( r _!) = k > 1S0 = i5i = i5( r ) = t, 2 So = 2 S\ is the highest in Stage II) 

Jfc-1 r-l (m-2)-(r-l) 

= J^ Yl {P(iS = iS l = i,(r-l-l) 1 S/s>k,l 1 S J 's = k, 

j=0 /=1 h=l 

h iSj s = i, (m — r — h) \Sj's < t, 2 So or 2 S\ is uniquely highest in Stage II) 
+ F(i5 = 1S1 = I, (r - 1 - /) i5/« > *, 

/ i5j's = k, h iS/s = i, (to - r - /1) i5/« < i, 2 5 = 2 Si is the highest in Stage II)} 



57 



Table 3.1. The joint distribution of IBD score of the locus 1 and locus 2. 



IBD of locus 1 


IBD of locus 2 


2 1 


2 

1 



$2 *(l-lf) (l-*) 2 
4 2 4 

>J(l-tf) * 2 + (l-*) 2 *(1-<I>) 
2 2 2 

(1-tf) 2 tf(l-tf) *2 
4 2 4 



Where 6 2 + (1 - B) 2 and is the recombination fraction between two loci. 



fc-1 r-l (m-2)-(r-l) 

E E E i p ^ = i^ = o t^t 



m-2! 



,=o ,=i h =i ( r ~ 1 - 0! « W ((m - 2) - (r- 1) - h)\ 

PdSj > k) r - l - ! PdSj = k) 1 P^Sj = z)' 1 PdSj < ir -2-(r-i)-h 



N 7 



J](2P( 2 5o = t, 2 Si <t) + P( 2 S = t, 2 Si = t)) x P{ 2 Sj < t) r - 1+h 



I t=i 



where A^2 = 



N — n\m 



r + l + h 



Int 



The joint distribution of the IBD scores of two markers (or genes), P(IBD of the 
marker (gene) 1, IBD of marker (gene) 2) is given in Table 3.1 adapted from Table 
2.3 where $=6 2 + (1 - 9) 2 and 6 is the recombination fraction between two loci. 

From Table 3.1, we can deduce the joint distribution of X {j and X x>1 for j ^ *'. 
The joint distribution between X {j and X {lj is shown in Table 3.2, with * defined as 
the # in Table 3.1. 

If a marker is not linked with the disease gene, then the probabilities of getting 
an IBD score of the marker equal to 2, 1, and are 0.25, 0.5, and 0.25, respectively. 
Thus, Xij has a Bernoulli distribution with parameter of 0.25. The distribution of ,Si 
under the null hypotheses is Binomial(n,, 0.25) for 1=1, 2. Let e be the probability 
that the disease is caused by the gene, and 9 be the distance between marker and the 
gene. For equations 3.5-3.7, where the recessive gene is at the end, X oj is distributed 



58 

as Bernoulli(e # 2 + (l-e) 0.25) and ,S as Binomial(n,, e V 2 + {l-e) 0.25). Similarly, 
for a dominant disease gene, X 0j is distributed as Bernoulli(e $/2 + (1 — e) 0.25) and 
iSi as Binomial(n,, e */2 + (1 - e) 0.25). 

For Eq. 3.11-3.13, we need to find the joint distribution P(/So=i, iSi—j), 
i=0,l,...,n/, j=0,l,...,n;, 1—1,2. Let n a &/ be the number of (Xoj, Xij) = (a, 6) in stage 
/. Then (n o;, "oi/, ^io/ 5 ini) has a multinomial distribution (n/, poo> Poi 5 Pw, Pu)- 
Thus, the joint distribution 



P{lS = i, f5i= j) 

= P(n 10 i + n u; =i, Hon + l»ui=j) 

min(i,j ) 

= jjP P(nm=k, n 10i =i - k, n ou =j - k, 

k = max(0, i + j — n) 

n m =n - i - j + nm), 



(3.27) 
(3.28) 



(3.29) 



can be computed once p o, Poi, Pio, and pn are specified. 

Although we assume the gene is in the middle of the two adjacent markers, the 
formulae are derived for the gene anywhere in between. Let the recombination fraction 
between the gene and marker be 6 Q and between the gene and marker 1 be 9\. Let 
62 be the recombination fraction between two markers. The joint distribution of X j 
and Xij, for x = 0, 1, y = 0, 1, is, 



Table 3.2. The joint distribution of X,j and X,/ 



Xij 


Xffj 


1 


1 




*2 (1-tf 2 ) 
4 4 
(1-tf 2 ) <p2 +2 
4 4 



59 



P(X oj = x, X X j = y) (3.30) 

= P{X 0j = x, Xij - y | gene IBD=2)P(gene IBD=2) 
+ P{X oj = x, X u = y | gene IBD=l)P(gene IBD=1) 
+ P(X 0j = x, Xij = y | gene elsewhere)P(gene elsewhere) (3.31) 

= P(X oj = x | X u ■ = y , gene IBD=2)P(X lj = y \ gene IBD=2) 
P(gene IBD=2) 
+ P(X j = x | Xxj = y, gene IBD=l)P(X lj = y \ gene IBD=1) 

P(gene IBD=1) 
+ P(X 0j = x | X\j = y, gene elsewhere) 

P(Xij = y | gene elsewhere)P( gene elsewhere). (3.32) 

Since an individual has to have two recessive disease genes in order to be affected, 
Eq. (3.32) becomes 

= P{X 0j = x | X u = y, gene IBD=2)P(* lj = y | gene IBD=2)e 

+ P(X 0j = x | Xij = y, gene elsewhere) P{Xi, = y \ gene elsewhere)(l - e). 

For a dominant disease, if an ASP is caused by a gene, then both sibs must at least 
share the disease gene and there is a 50-50 chance they share the other allele. Thus, 
Eq. (3.32) becomes 

= P(X oj = x | Xu = y, gene IBD=2)P(A' lj = y \ gene IBD=2)(e/2) 
+ P(X oj = x | Xu = y, gene IBD=l)P(X li = y | gene IBD=l)(e/2) 
+ P(X j = x | Xij = y, gene elsewhere) 

^(^ij = V I gene elsewhere)(l - e). (3.33) 

Let f |. 8? + (1 - 9{)\ /=0,1,2, p xy = P(X oi = x | X h = y), x « 0,1, y . 0,1. Then 



60 



apply Table 3.2 for a recessive disease we have 

Poo = (l-*g)(l-*f) £ + JS_t2(l-c), (3.34) 

Pio = tKl-t?)e + i^.(l-«), (3-35) 

Poi = (1-«5)«J « + -^-(l-«), (3-36) 

Pn = *l*l e + ^(l-e). (3.37) 
For a dominant disease, 

Poo = (1-«8)(1-«J) | + (l-»o+t8)(l-ti+f?)|+-i±i(l-e) 1 (3.38) 



Pl0 = ^(l-«?)|+'lo(^«o)(l-ti+«?)| + -^^(l-«), (3-39) 

poi = (l-t2)*;e + (l-t.+»J)t 1 (l-»i)| + ii^2-(],-«), (3-40) 

pi, - *: *i | + »o<i-*o)»i(i-«i) | + ^ (i - «)- (3- 41 ) 

Thus, P(the marker closest to the gene is found) can be calculated. In addition to 
analysis computation for Eq. 3.3 and 3.4, simulations were also done. 

3.3.2 Simulation under More Realistic Assumptions 

The assumption of independent Xy violates the fact that some loci are linked. 
Simulations were done under a Markov chain model with the combinations of resource 
N = 1000, 2000, 5000, and 10,000, number of ASP n = 10, 25, 75, and 100, e = 0.25, 
0.5, 0.75, and 1, m from 50 to 350 with an increment 25 for two-stage design and 10 
for one-stage design, r from 5 to m with an increment (m — 5)/10, and ni from 5 to 

n — 5 with increment 5. 

The simulations were conducted as follows: 



61 



1. Reading in parameters, resource N, heterogeneity s, number of ASP n, and m, 
ill, and r of the design. 

2. Given ni, m, generate y from uniform(0,l). If y < e then generate gene location 
from uniform distribution (0,3300). Let locus / and locus / + 1 be two adjacent 

loci, i.e. the gene location is between P* and » Zj , where L is the total length 
of genome. This step determines which interval contains the gene and then: 

(a) For recessive disease, generate IBD scores at loci / and / + 1 conditional 
on both sibs of ASP carrying two disease genes and the gene locus is in 
the middle of two markers. Haldane map function is used to convert map 
distance into recombination fraction 9. 

(b) For dominant disease, if y < 0.5e, generate IBD scores at loci / and / + 1 
conditional on both sibs of ASP sharing two alleles at gene loci and the 
gene locus is in the middle of two markers. If 0.5e < y < e, then generate 
IBD score at loci / and / + 1 conditional on both sibs of ASP share one allele 
at gene loci and, again, the gene locus is in the middle of two markers. 

If y > e, let I =0, and generate IBD score at locus from Bernoulli(0.25) as if 
there is no gene linked with marker 0. 

3. For Markov chain model simulation, generate IBD score at locus i, i start from 
/ — 1 and decrease to 0, conditional on IBD score at locus i-f-1, and then generate 
IBD score at locus i conditional on IBD score at locus i — 1 for i — I + 1, ..., 
m — 1. For independent model simulation, generate IBD scores independently 
from Trinomial(rci, 0.25, 0.5,0.25). Conditional probability formulas are given 
by Table 2.3. 

4. Convert IBD scores into statistics X-,s. 

5. Repeat step 2, 3, and 4 n x times and then calculated iS{. 



62 



6. Check ties, adjust R to include all the ties with the rth highest \S{ for Stage 
II, check whether \Sq passes Stage I. If yes, then calculate A^ according to the 
resource constraint and go to the next step. If not, record the detection as a 
failure and go back to step 2. 

7. The XijS in Stage II are generated the same way as those in Stage I, except 
only the chosen markers are used and repeat N 2 times. 

8. Check whether 250 or 2S1 is the unique largest among 2 SiS in Stage II. 

9. If marker or 1 is chosen, it is a success; otherwise it is a failure. Go back to 
step 2 until enough simulation has been done. 

The programs were compiled using GCC version 2.7.2 on a PC with Pentium 166 
CPU and 48M RAM in an OS/2 environment. The random number generator for the 
simulation was adapted from Press (1992). 

3.3.3 Results 

Based on analytic computation and simulation, the designs with the highest prob- 
ability of finding the right marker (power) were identified. These optimal designs are 
given in Table 3.3 for searching a recessive gene, and Table 3.4 for a dominant gene. In 
both tables, in the ASP column is reported the number of available affected-sib-pairs; 
the e column shows the probability that the disease is caused by the gene, repre- 
senting heterogeneity; the m columns, the number of marker loci used in the first 
stage; the r column, the proposed number of loci to be chosen in Stage I for Stage 
II study; the n x columns, the number of ASP used in first stage; the F2 column, 
the probability of locating the right marker by the best two-stage design obtained by 
analytic formula; the Indep columns and the Markov columns show the probabilities 
obtained by simulation with independent assumption and Markov chain assumption 
without combining first- and second-stage data; and the Comb, column shows the 



63 



simulated probability of the two-stage design with Markov chain model with first- 
and second-stage data combined. The Fl column gives the probability of the best 
one-stage design, i.e., with optimal m and n subject to mn < N obtained by analytic 
formula. The last column, F2— Fl, shows the increase in probability of two-stage de- 
sign over one-stage. An asterisk is marked when the increase was over 0.35 for Table 
3.3 (recessive) and 0.15 for Table 3.4 (dominant). 



64 



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in io io io 


IO ° « iO 


w io ° ° 

rH "" rH rH 


^ o o m 

J rH rH rH 


m o m o 

rH rH rH CS 


— 


K5 iO iO lO 


m m N w 

J CS CO 


.- CO C7> C75 

*" ■* cs cs 


O CN Tf Tf 

■* cs cs cs 


CO O) CN N 
CS rH CM rH 


£ 


C-j _ O IO 


— , iO m o 
S CS CS O 

10 H N M 


a o o o 
a o m m 

10 CN CN CN 


^ m o o 

' rH CS CS 


o o m in 

o m N N 

r^ r—i i—i r—i 




0J 

H - 


io 


in O iO O 
CS IO t-- o 

d d d r* 


m o m o 
cs io r- o 

d d d i-h 


m o m o 
CN in t-- o 

O O O rH 


m o m o 
N in N o 
d d d ^ 


to o m o 
cs in c- o 

d d d ^i 


(13 

1* 


2h 

< 


o o o o 

i — 1 i— ( i— 4 i— I 


lO IO IO iO 
CN CS CS CS 


o o o o 
m m m m 


in m in m 
t— t- t- t- 


o o o o 

o o o o 

^ r~t i — 1 rH 



71 



-a 
o> 

b 

.2 

c 
o 
o 
I 

CO 
-O 





> 


r-H 


o 
a 


o 


CN 


r- 


Ol 


•* 


f~ 


o 


!—~ 


no 


CO 


no 


CN 


CO 


m 


r^ 


CN 


r^ 


a> co 




o 


fa 


o 


o 


o 


o 


— i 


■<t 


r~ 


i-H 


CN 


'.N 


a 


o 


^r 


CO 




■* 


r-H O 






o 


o 


o 


co 


o 


CO 


o 


O 


o 


o 


C_> 


a 


O 


o 


o 


' — 


o 


T-H r-H 




CM 

fa 


o 


o 

1 


o 


o 


O 


o 


CO 
1 


o 

1 


CO 


o 


o 


O 


CO 


o 


o 


o 


CO 
1 


o 


o o 






> 

n 


on 


en 


■*f 


CO 


— i 


r— 1 


so 


o 


o 


^p 


CO 


r-H 


o 


CO 


^r 


^r 


o 


<ji 


CN t}< 






^ 


IN 


i-n 


in 


on 


Tf 


r- 


T 


r» 


m 


■* 


m 


a> 


u; 


r-H 


CO 


CN 


oo 


"j 


CN 00 






— 
rt 

s 


a 


o 


o 


O 


o 


o 


r-H 


CN 


CO 


1— t 


CO 


m 


CO 


CN 


^P 


l^ 


o 


CN 


m r— 




a 
bO 


o 


o 


o 


O 


CO 


o 


o 


o 


O 


o 


o 


o 


o 


O 


O 


O 


CO 


O 


o o 


n, 


D 


on 


r^ 


<~> 


on 


CN 


CN 


CD 


CO 


i-H 


CO 


o> 


r~ 


r- 


o 


o 


in 


f- 


r-H in 




w 


yj 


nn 


in 


m 


no 


c~ 


!0 


r-H 


m 


CO 


00 


CO 


EC 


CN 


C75 


Tf 


oo 


CO 


■^ 00 




CD 

-o 

CV 
bO 


-a 


CO 


O 


O 


O 


o 


o 


i— I 


CO 


CO 


T-H 


CO 


CO 


CO 


CN 


■* 


(^ 


o 


CN 


m f~ 




B 


o 


O 


O 


O 


o 


o 


o 


o 


O 


o 


o 


o 


o 


O 


O 


o 


CO 


o 


o o 




on 


h- 


<"N 


05 


o 


05 


00 


m 


CN 


no 


00 


o 


•>* 


00 


f- 


CD 


i — i 


m 


CO o 






rs 


m 


m 


on 


•^f 


CO 


iO 


r-H 


m 


m 


CO 


CO 


ep 


r-H 


t^ 


CO 


X 


in 


CN 00 




Cfi 


fa 


O 


o 


O 


O 


CO 


O 


r-H 


no 


o 


r-H 


CO 


CO 


CO 


CN 


■<* 


h- 


CO 


CN 


m r- 




i 



O 


O 


o 


O 


CO 


CO 


o 


a 


O 


o 


o 


o 


o 


o 


o 


o 


o 


CO 


o 


o o 




o 


O 


e 


o 


lO 


in 


m 


m 


CO 


o 


o 


o 


m 


m 


lO 


r-H 


CO 
CO 
1 — 1 


o 
o 

t-H 


o o 
o o 

T-H T-H 








I— 1 


r-H 


r-H 


r-H 


CN 


CN 


<N 


CN 


35 


m 


in 


m 


t- 


r- 


f- 


t^ 








(—1 


Q 




Q 


Q 


o 


o 
35 


Q 


o 


o 


r~. 


o 


o 


o 


CO 


o 


o o 






6 




< 1 


m 

1— 1 


CN 


m 


oo 

T-H 


CO 


m 
no 


O 
CN 


o 

CN 


o 

CN 


co 


CO 


CO 


T}< 


CO 


o 


o o 


o 
o 






















































































o 




o 


on 


CN 


h- 


r- 


CO 


oo 


■* 


o 


00 


■*r 


■<* 


T-H 


00 


00 


Tf 


m 


o 


m 


CN C5 


o 




E 

o 

Q 


<N 


•*r 


m 


C5 


—■ 


CO 


-T 


h> 


bQ 


"* 


■* 


o> 


m 


CN 


1 — 1 


t- 


1- 


oo 


cn m 


1—1 




O 


o 


o 


O 


co 


o 


1— 1 


CN 


O 


t-H 


CO 


m 


CO 


CN 


m 


t-~ 


co 


CN 


CO 00 


II 
2 




O 


o 


o 


o 


CO 


o 


CO 


O 


e 


o 


o 


o 


CO 


o 


o 


o 


o 


O 


o o 


y 




> 




r~ 


o 


r-n 


CO 


re 


no 


~T 


^r 


— 


Tf 


a> 


o 


t- 


m 


CO 


T-H 


00 


T-H 


t}< m 


3 




_* 


IN 


■«t 


in 


on 


~r 


CO 


co 


■* 


~T 


CO 


T-H 


m 


-* 


o 


N- 


no 


-r 


CO 


CT) CN 


Q 

co 

0! 




l_ 


o 


o 


o 


O 


CO 


o 


1 — 1 


CN 


CO 


T-H 


no 


m 


CO 


CN 


Tf 


t— 


CO 


CN 


m oo 


B 
bO 


o 


o 


o 


O 


o 


o 


CO 


o 


CO 


o 


O 


o 


o 


O 


o 


O 


e 


o 


o o 














































en 


a. 


— 1 


on 


■*f 


t-H 


es 


CO 


0) 


■* 


1- 


m 


r-H 


m 


\~ 


r- 


00 


CO 


CO 


00 


r-~ co 




•o 


0) 


CO 


no 


in 


oo 


— 


CO 


TO 


r~ 


" 


Tf 


m 


o 


— 


r—t 


CN 


o> 


It 


CO 


m t-- 




t> 


o 


o 


o 


o 


CO 


O 


— 1 


CN 


o 


t-H 


no 


CO 


CO 


CN 


in 


r~ 


o 


CN 


CO 00 




QJ 
bC 

73 


s 


o 


o 


o 


o 


CO 


o 


CO 


o 


CO 


o 


o 


o 


CO 


O 


o 


o 


CO 


o 


O o 




on 


CO 


Q 


ON 


N 


CO 


— 1 


m 


n 


o 


m 


r~ 


n 


■* 


CN 


CN 


cr: 


CO 


m co 




o 


CN 


(N 


CO 


in 


on 


-p 


CO 


Tf 


r~ 


its 


rf 


t* 


o 


te 


CN 


CN 


O 


I- 


o 


■* 00 




a 


o 


O 


o 


CO 


o 


— H 


CN 


CO' 


t-H 


CO 


CO 


CO 


CN 


m 


00 


co 


CO 


CO 00 






o 


o 


O 


o 


o 


o 


o 


o 


o 


o 


o 


o 


CO 


o 


o 


o 


CO 


o 


o o 


B 


It 


m 


m 


in 


It 


o 

T-H 


It 


m 


Ifl 

— 1 


m 


iC 


m 


m 


o 

T-H 


lO 

T-H 


m 

r— t 


it 


lO 

< — t 


m m 

r-H CN 


E-H 


LC 


IO 


m 


m 


in 


m 


1- 

CN 


CO 

CO 


iC 


CO 

05 


— 1 

CO 

i — t 


m 

CN 

T-H 




r-H 


as 

m 


m 


>t 
cC 


OS 


OJ C75 
Tf CN 




Q 

It 


o 


a 


in 


CO 

it 


in 


lO 


o 


o 


in 


o 


o 


CO 


m 


m 


m 


Lt 


m 


m o 






E 


© 


CN 


CN 


CN 


o 


CN 


m 


m 


CO 


CN 


i-~ 


r^ 


OJ 


CN 


cn m 






t— i 


CN 


i— 1 


CN 


no 


CN 


CN 


CN 


1 — 1 


CN 


CN 


CN 


i « 


CN 


CN CN 






It 


o 


in 


O 


Ifl 


O 


m 


o 


m 


O 


m 


O 


kA 


O 


m 


o 


It 


O 


in o 




tu 


III 


•M 


IO 


r- 


o 


w 


m 


t-- 


o 


N 


m 


r-~ 


o 


ss 


m 


i-- 


o 


CN 


m 


t~ o 




0) 

S 

— 
fa 




o 


o 


o 


1 — 1 


o 


o 


o 


i— i 


CO 


o 


o 


r-H 


CO 


o 


o 


r-H 


CO 


o 


O r-H 


fa 

in 
< 


n 


o 


o 


o 


It 


m 


m 


m 


o 


o 


o 


O 


lO 


m 


lO 


in 


CO 
CO 
— 1 


o 

o 

T-H 


O O 

o o 

r-H r-H 




- — i 


1 — 1 


r-H 


1 — 1 


w 


CN 


CM 


CN 


35 


m 


m 


m 


t- 


o 


t- 


t~ 



72 



3.4 Type I Error and Power of Claiming Linkage 

Tables 3.3 and 3.4 give the optimal designs and the probabilities of finding the 
locus linked to the responsible gene when the gene exists, but they do not provide 
information on the probability of causing a false conclusion when there is no gene 
responsible for the disease. The usual requirement is that the LOD score in Stage II 
should be greater than a certain threshold, t, in order to claim linkage. Given R = r, 
we let the threshold be t, which is the lOO(l-a) percentile of the unique maximum, 
T, of r Binomial(n 2 , 0.25) random variables. The probability mass function of T is 

P(T = s) 

= P(2Si Q = s\ 2 Si is the unique maximum) 

P( 2 Si = s, 2 Si is the unique maximum) 
P( 2 Si is the unique maximum) 

rf(s)F(s-l) r - 1 



I&rmFia-lY-* 

The probability mass function of the marker group is 

P(T = s) 

= P{ 2 Si = 2 Si 0+ i = 5 ^/o and 2 57 are the unique maximum) 

P( 2 Si = 2 Si +i — s, 2 Si Q \s the unique maximum) 
P{ 2 Si and 25/0+1 are the unique maximum) 



r-2 



(r-l)f(s)*F{8-l) 

ZZAr-i)mF(s-iy-^ 

Where f(s) and F(s) are probability mass function and cumulative distribution 
function of Binomial(n 2 , 0.25). Table 3.5 and 3.6 gives the value of t of the unique 
maximum and the marker group. To use Table 3.5 and 3.6, first find your n 2 in the 
n 2 column, then find your R in R column, if your R is not in the table, then find the 



73 



largest value R that is smaller than your R with same n 2 . The value in the t column 
is the 95% percentile. The range of n 2 is from 5 to 95, and for R is from 5 to 300. 

A linkage can be claimed if and only if 2 S/ > t, where I is one of the markers 
that was chosen in Stage II. Clearly, 

P(linkage claim is incorrect) 

< P(2Si Q > t | no gene is responsible) jP(no gene is responsible) 
+P(I is wrong and 2 S7 > * | a gene is responsible) 
P(a gene is responsible). 

The (prior) P(no gene is responsible) is usually unknown, but we can conclude that 
P(/ is wrong and 2 S/ >t\& gene is responsible) < P( 2 S/ > * I no gene is respon- 
sible). A proof is as follows. Let A denote the event { the gene is between marker i 
and marker i + 1}. 

P(I is wrong and 2 Si > t \ a. gene is responsible) (3-42) 

= P{ 2 S Io > 2 Sj, Vj # /„, 2 S Io > t and /„ / i ovio + l \A) (3.43) 

= Y, P(* = 25/ > 2 Sj, Vj ^ /<,, and /o + i or i + 1 \A) (3.44) 

k=t+l 
n 2 



= Y P(k > a^ii V? ^ ^°'* ' or 'o + 1» * > aSfei * > 25 , l0+ i, 

fcst+1 

2 5 7o = fc and /„ / to or i + 1 |A) (3.45) 

= ^ {P(fc > ,5y, Vj ^ /o, H>, or i + 1, / + i or i + 1 \A) (3.46) 

P(k > 2 S l0 , k > 2 S t0+l \A) (3.47) 

^(25/0 = ■*, /o 9^ ^o or i + 1 |A)>, (3.48) 



74 



Table 3.5. The 95% percentile of the unique maximum of R Binomial(n 2 , 0.25). 



n 2 


R 


1 


"2 


R 


t 


n 2 


R 


t 


n 2 


R 


( 


m j 


R 


t 


"2 


R 


( 


5 


5 


5 


28 


44 


16 


45 


20 


21 


59 


145 


28 


72 


255 


33 


84 


266 


37 


5 


24 


5* 


28 


149 


17 


45 


46 


22 


60 


5 


23 


73 


5 


27 


85 


5 


30 


6 


5 


5 


29 


5 


14 


45 


120 


23 


60 


6 


24 


73 


7 


28 


85 


6 


31 


6 


7 


1 


29 


11 


15 


46 


5 


19 


60 


12 


25 


73 


12 


29 


85 


9 


32 


6 


84 


6* 


29 


29 


16 


46 


7 


20 


60 


22 


26 


73 


22 


30 


85 


15 


33 


7 


5 


6 


29 


92 


17 


46 


15 


21 


60 


48 


27 


73 


42 


31 


85 


26 


34 


7 


22 


7 


30 


5 


14 


46 


34 


22 


60 


108 


28 


73 


88 


32 


85 


50 


35 


8 


5 


6 


30 


8 


15 


46 


85 


23 


60 


264 


29 


73 


193 


33 


85 


99 


36 


8 


9 


7 


30 


20 


16 


46 


233 


24 


61 


5 


24 


74 


5 


27 


85 


207 


37 


8 


69 


8 


30 


60 


17 


47 


5 


19 


61 


10 


25 


74 


6 


28 


86 


5 


31 


9 


5 


7 


30 


202 


18 


47 


6 


20 


61 


18 


26 


74 


10 


29 


86 


8 


32 


9 


24 


8 


31 


5 


14 


47 


12 


21 


61 


37 


27 


74 


18 


30 


86 


13 


33 


9 


225 


9 


31 


1 


15 


47 


26 


22 


61 


82 


28 


74 


34 


31 


86 


22 


34 


10 


5 


7 


31 


15 


16 


47 


62 


23 


61 


194 


29 


74 


69 


32 


86 


41 


35 


10 


11 


8 


31 


40 


17 


47 


162 


24 


62 


5 


24 


74 


148 


33 


86 


79 


36 


10 


70 


9 


31 


126 


18 


48 


5 


20 


62 


8 


25 


75 


5 


28 


86 


162 


37 


11 


5 


7 


32 


5 


15 


48 


10 


21 


62 


15 


26 


75 


9 


29 


87 


5 


31 


11 


6 


8 


32 


11 


16 


48 


20 


22 


62 


29 


27 


75 


15 


30 


87 


7 


32 


11 


29 


9 


32 


28 


17 


48 


46 


23 


62 


63 


28 


75 


28 


31 


87 


11 


33 


11 


216 


10 


32 


81 


18 


48 


115 


24 


62 


144 


29 


75 


55 


32 


87 


18 


34 


12 


5 


8 


32 


273 


19 


49 


5 


20 


63 


5 


24 


75 


115 


33 


87 


33 


35 


12 


14 


9 


33 


5 


15 


49 


8 


21 


63 


7 


25 


75 


256 


34 


87 


64 


36 


12 


79 


10 


33 


8 


16 


49 


16 


22 


63 


12 


26 


76 


5 


28 


87 


128 


37 


13 


5 


8 


33 


20 


17 


49 


34 


23 


63 


23 


27 


76 


7 


29 


87 


271 


38 


13 


8 


9 


33 


54 


18 


49 


83 


24 


63 


49 


28 


76 


13 


30 


88 


5 


31 


13 


36 


10 


33 


171 


19 


49 


219 


25 


63 


108 


29 


76 


23 


31 


88 


6 


32 


13 


231 


11 


34 


5 


15 


50 


5 


20 


63 


259 


30 


76 


44 


32 


88 


9 


33 


14 


5 


9 


34 


6 


16 


50 


6 


21 


64 


5 


24 


76 


90 


33 


88 


16 


34 


14 


19 


10 


34 


14 


17 


50 


12 


22 


64 


6 


25 


76 


196 


34 


88 


28 


35 


14 


95 


11 


34 


37 


18 


50 


26 


23 


64 


10 


26 


77 


5 


28 


88 


52 


36 


15 


5 


9 


34 


111 


19 


50 


61 


24 


64 


19 


27 


77 


6 


29 


88 


102 


37 


15 


11 


10 


35 


5 


16 


50 


155 


25 


64 


38 


28 


77 


11 


30 


88 


212 


38 


15 


46 


11 


35 


11 


17 


51 


5 


21 


64 


83 


29 


77 


19 


31 


89 


5 


32 


15 


265 


12 


35 


27 


18 


51 


10 


22 


64 


192 


30 


77 


35 


32 


89 


8 


33 


16 


5 


9 


35 


74 


19 


51 


20 


23 


65 


5 


25 


77 


71 


33 


89 


13 


34 


16 


7 


10 


35 


232 


20 


51 


46 


24 


65 


8 


26 


77 


151 


34 


89 


23 


35 


16 


25 


11 


36 


5 


16 


51 


112 


25 


65 


15 


27 


78 


5 


28 


89 


43 


36 


16 


119 


12 


36 


8 


17 


51 


296 


26 


65 


30 


28 


78 


6 


29 


89 


82 


37 


17 


5 


10 


36 


19 


18 


52 


5 


21 


65 


64 


29 


78 


9 


30 


89 


167 


38 


17 


15 


11 


36 


51 


19 


52 


8 


22 


65 


144 


30 


78 


16 


31 


90 


5 


32 


17 


60 


12 


36 


150 


20 


52 


16 


23 


66 


5 


25 


78 


29 


32 


90 


7 


33 


18 


5 


10 


37 


5 


16 


52 


35 


24 


66 


7 


26 


78 


57 


33 


90 


11 


34 


18 


9 


11 


37 


7 


17 


52 


82 


25 


66 


13 


27 


78 


118 


34 


90 


20 


35 


18 


33 


12 


37 


14 


18 


52 


209 


26 


66 


24 


28 


78 


258 


35 


90 


35 


36 


18 


152 


13 


37 


36 


19 


53 


5 


21 


66 


50 


29 


79 


5 


29 


90 


67 


37 


19 


5 


10 


37 


100 


20 


53 


7 


22 


66 


109 


30 


79 


8 


30 


90 


133 


38 


19 


6 


11 


38 


5 


17 


53 


13 


23 


66 


256 


31 


79 


13 


31 


90 


277 


39 


19 


20 


12 


38 


11 


18 


53 


27 


24 


67 


5 


25 


79 


24 


32 


91 


5 


32 


19 


79 


13 


38 


26 


19 


53 


61 


25 


67 


6 


26 


79 


46 


33 


91 


6 


33 


20 


5 


11 


38 


69 


20 


53 


150 


26 


67 


11 


27 


79 


93 


34 


91 


10 


34 


20 


13 


12 


38 


205 


21 


54 


5 


21 


67 


20 


28 


79 


199 


35 


91 


17 


35 


20 


45 


13 


39 


5 


17 


54 


6 


22 


67 


39 


29 


80 


5 


29 


91 


29 


36 


20 


196 


14 


39 


9 


18 


54 


10 


23 


67 


84 


30 


80 


7 


30 


91 


54 


37 


21 


5 


11 


39 


19 


19 


54 


21 


24 


67 


192 


31 


80 


11 


31 


91 


106 


38 


21 


9 


12 


39 


48 


20 


54 


46 


25 


68 


5 


26 


80 


20 


32 


91 


217 


39 


21 


27 


13 


39 


136 


21 


54 


110 


26 


68 


9 


27 


80 


37 


33 


92 


5 


32 


21 


105 


14 


40 


5 


17 


54 


282 


27 


68 


16 


28 


80 


74 


34 


92 


6 


33 


22 


5 


11 


40 


7 


18 


55 


5 


22 


68 


31 


29 


80 


154 


35 


92 


9 


34 


22 


6 


12 


40 


15 


19 


55 


9 


23 


68 


65 


30 


81 


5 


29 


92 


14 


35 


22 


17 


13 


40 


35 


20 


55 


17 


24 


68 


145 


31 


81 


6 


30 


92 


24 


36 


22 


60 


14 


40 


93 


21 


55 


35 


25 


69 


5 


26 


81 


10 


31 


92 


45 


37 


22 


256 


15 


40 


278 


22 


55 


81 


26 


69 


8 


27 


61 


17 


32 


92 


86 


38 


23 


5 


12 


41 


5 


17 


55 


202 


27 


69 


13 


28 


81 


30 


33 


92 


172 


39 


23 


12 


13 


41 


6 


18 


56 


5 


22 


69 


25 


29 


81 


59 


34 


93 


5 


33 


23 


37 


14 


41 


11 


19 


56 


7 


23 


69 


51 


30 


81 


121 


35 


93 


8 


34 


23 


140 


15 


41 


26 


20 


56 


13 


24 


69 


111 


31 


81 


262 


36 


93 


12 


35 


24 


5 


12 


41 


65 


21 


56 


28 


25 


69 


255 


32 


82 


5 


30 


93 


21 


36 


24 


8 


13 


41 


185 


22 


56 


61 


26 


70 


5 


26 


82 


8 


31 


93 


37 


37 


24 


24 


14 


42 


5 


18 


56 


147 


27 


70 


7 


27 


82 


14 


32 


93 


70 


38 


24 


82 


15 


42 


9 


19 


57 


5 


22 


70 


11 


28 


82 


25 


33 


93 


137 


39 


25 


5 


12 


42 


19 


20 


57 


6 


23 


70 


21 


29 


82 


48 


34 


93 


284 


40 


25 


6 


13 


42 


47 


21 


57 


11 


24 


70 


41 


30 


82 


96 


35 


94 


5 


33 


25 


16 


14 


42 


127 


22 


57 


22 


25 


70 


86 


31 


82 


202 


36 


94 


7 


34 


25 


50 


15 


43 


5 


18 


57 


47 


26 


70 


192 


32 


83 


5 


30 


94 


11 


35 


25 


188 


16 


43 


7 


19 


57 


108 


27 


71 


5 


26 


83 


7 


31 


94 


18 


36 


26 


5 


13 


43 


15 


20 


57 


272 


28 


71 


6 


27 


83 


12 


32 


94 


31 


37 


26 


11 


14 


43 


34 


21 


58 


5 


23 


71 


9 


28 


83 


21 


33 


94 


57 


38 


26 


32 


15 


43 


89 


22 


58 


9 


24 


71 


17 


29 


83 


39 


34 


94 


110 


39 


26 


110 


16 


43 


252 


23 


58 


17 


25 


71 


33 


30 


83 


76 


35 


94 


224 


40 


27 


5 


13 


44 


5 


18 


58 


36 


26 


71 


67 


31 


83 


158 


36 


95 


5 


33 


27 


8 


14 


44 


6 


19 


58 


81 


27 


71 


146 


32 


84 


5 


30 


95 


6 


34 


27 


22 


15 


44 


12 


20 


58 


197 


28 


72 


5 


27 


84 


6 


31 


95 


9 


35 


27 


68 


16 


44 


26 


21 


59 


5 


23 


72 


8 


28 


84 


10 


32 


95 


15 


36 


27 


251 


17 


44 


63 


22 


59 


8 


24 


72 


14 


29 


84 


18 


33 


95 


26 


37 


28 


5 


13 


44 


172 


23 


59 


14 


25 


72 


26 


30 


84 


32 


34 


95 


47 


38 


28 


6 


14 


45 


5 


19 


59 


28 


26 


72 


53 


31 


84 


61 


35 


95 


89 


39 


28 


15 


15 


45 


9 


20 


59 


62 


27 


72 


113 


32 


84 


124 


36 


95 


178 


40 



75 



Table 3.6. The 95% percentile of the unique maximum marker group of R 
Binomial(n 2 , 0.25). 



"2 


ft 


t 


"2 


R 


t 


n? 


ft 


t 


n 2 


ft 


t 


n 2 


ft 


t 


n. 


Ft 


t 


1 


5 


4 


2« 


64 


15 


u 


8 


18 


59 


108 


26 


li 


9 5 


30 


84 


115 


34 


5 


12 


■ 


28 


203 


16 


45 


16 


19 


59 


248 


27 


72 


201 


31 


84 


227 


35 


5 


173 


5* 


29 


5 


12 


45 


33 


20 


60 


5 


21 


73 


I 


25 


85 


5 


28 


6 


5 


4 


29 


8 


13 


45 


76 


21 


60 


7 


22 


73 


8 


26 


85 


6 


29 


6 


6 


5 


29 


17 


14 


45 


193 


22 


60 


11 


23 


73 


12 


27 


85 


10 


30 


6 


38 


6 


29 


44 


15 


46 


5 


17 


60 


20 


24 


73 


21 


28 


85 


16 


31 


7 


5 


5 


29 


131 


16 


46 


7 


18 


60 


40 


25 


73 


39 


29 


85 


27 


32 


7 


16 


6 


30 


5 


12 


46 


13 


19 


60 


83 


26 


73 


76 


30 


85 


49 


33 


7 


132 


7 


30 


6 


13 


46 


26 


20 


60 


187 


27 


73 


158 


31 


85 


92 


34 


8 


5 


5 


30 


13 


14 


46 


57 


21 


61 


5 


21 


74 


5 


25 


85 


181 


35 


8 


8 


6 


30 


31 


15 


46 


140 


22 


61 


6 


22 


74 


7 


26 


86 


5 


28 


8 


43 


7 


30 


87 


16 


47 


5 


17 


61 


10 


23 


74 


11 


27 


86 


6 


29 


9 


5 


6 


30 


278 


17 


47 


6 


18 


61 


17 


24 


74 


18 


28 


86 


9 


30 


9 


21 


7 


31 


5 


13 


47 


10 


19 


61 


32 


25 


74 


32 


29 


86 


14 


31 


9 


130 


8 


31 


10 


14 


47 


21 


20 


61 


65 


26 


74 


62 


30 


86 


23 


32 


10 


5 


6 


31 


23 


15 


47 


44 


21 


61 


143 


27 


74 


125 


31 


86 


41 


33 


10 


12 


7 


31 


60 


16 


47 


103 


22 


62 


5 


22 


74 


265 


32 


86 


75 


34 


10 


54 


8 


31 


179 


17 


47 


262 


23 


62 


8 


23 


75 


5 


25 


86 


146 


35 


11 


5 


6 


32 


I 


13 


48 


5 


18 


62 


14 


24 


75 


6 


26 


86 


296 


36 


11 


7 


7 


32 


8 


14 


48 


9 


19 


62 


26 


25 


75 


9 


27 


87 


5 


29 


11 


28 


8 


32 


17 


15 


48 


17 


20 


62 


52 


26 


75 


15 


28 


87 


8 


30 


11 


151 


9 


32 


43 


16 


48 


34 


21 


62 


110 


27 


75 


27 


29 


87 


12 


31 


12 


5 


7 


32 


120 


17 


48 


77 


22 


62 


249 


28 


75 


50 


30 


87 


20 


32 


12 


16 


8 


33 


5 


13 


48 


189 


23 


63 


5 


22 


75 


99 


31 


87 


34 


33 


12 


70 


9 


33 


7 


14 


49 


5 


18 


63 


7 


23 


75 


207 


32 


87 


62 


34 


13 


■ 


7 


33 


13 


15 


49 


7 


19 


63 


12 


24 


76 


5 


26 


87 


118 


35 


13 


10 


8 


33 


31 


16 


49 


13 


20 


63 


22 


25 


76 


8 


27 


87 


235 


36 


13 


37 


9 


33 


82 


17 


49 


27 


21 


63 


42 


26 


76 


13 


28 


88 


5 


29 


13 


187 


10 


33 


246 


18 


49 


58 


22 


63 


86 


27 


76 


23 


29 


88 


7 


30 


14 


5 


7 


34 


B 


14 


49 


138 


23 


63 


189 


28 


76 


41 


30 


88 


10 


31 


14 


7 


8 


34 


10 


15 


50 


5 


18 


64 


5 


22 


76 


80 


31 


88 


17 


32 


14 


22 


9 


34 


23 


16 


50 


6 


IS 


64 


6 


23 


76 


163 


32 


88 


29 


33 


14 


93 


10 


34 


58 


17 


50 


11 


20 


64 


10 


24 


77 


5 


26 


88 


51 


34 


15 


5 


8 


34 


164 


18 


50 


21 


21 


64 


18 


25 


77 


7 


27 


88 


96 


35 


15 


14 


9 


35 


5 


14 


50 


45 


22 


64 


34 


26 


77 


11 


28 


88 


188 


36 


15 


51 


10 


35 


8 


15 


50 


103 


23 


64 


68 


27 


77 


19 


29 


89 


5 


29 


15 


240 


11 


35 


18 


16 


50 


255 


24 


64 


146 


28 


77 


34 


30 


89 


6 


30 


16 


5 


8 


35 


42 


17 


51 


5 


19 


65 


5 


23 


77 


65 


31 


89 


9 


31 


16 


9 


9 


35 


113 


18 


51 


9 


20 


65 


9 


24 


77 


129 


32 


89 


15 


32 


16 


30 


10 


36 


5 


14 


51 


17 


21 


65 


15 


25 


77 


272 


33 


89 


24 


33 


16 


124 


11 


36 


7 


15 


51 


35 


22 


65 


28 


26 


78 


5 


26 


89 


43 


34 


17 


5 


8 


36 


14 


16 


51 


78 


23 


65 


54 


27 


78 


6 


27 


89 


79 


35 


17 


7 


9 


36 


31 


17 


51 


186 


24 


65 


113 


28 


78 


10 


28 


89 


152 


36 


17 


19 


10 


36 


79 


18 


52 


5 


19 


65 


252 


29 


78 


16 


29 


90 


5 


29 


17 


69 


11 


36 


225 


19 


52 


8 


20 


66 


5 


23 


78 


29 


30 


90 


6 


30 


18 


5 


9 


37 


5 


14 


52 


14 


21 


66 


8 


24 


78 


53 


31 


90 


8 


31 


18 


13 


10 


37 


6 


15 


52 


28 


22 


66 


13 


25 


78 


103 


32 


90 


13 


32 


18 


41 


11 


37 


11 


16 


52 


60 


23 


66 


23 


26 


78 


213 


33 


90 


21 


33 


18 


166 


12 


37 


24 


17 


52 


138 


24 


66 


44 


27 


79 


5 


26 


90 


36 


34 


19 


5 


9 


37 


57 


18 


53 


5 


19 


66 


89 


28 


79 


6 


27 


90 


65 


35 


19 


9 


10 


37 


154 


19 


53 


7 


20 


66 


193 


29 


79 


9 


28 


90 


123 


36 


19 


26 


11 


38 


5 


15 


53 


12 


21 


67 


5 


23 


79 


14 


29 


90 


244 


37 


19 


94 


12 


38 


9 


16 


53 


22 


22 


67 


7 


24 


79 


24 


30 


91 


5 


30 


20 


5 


9 


38 


18 


17 


53 


47 


23 


67 


11 


25 


79 


44 


31 


91 


7 


31 


20 


7 


10 


38 


42 


18 


53 


104 


24 


67 


19 


26 


79 


83 


32 


91 


11 


32 


20 


18 


11 


38 


108 


19 


53 


251 


25 


67 


35 


27 


79 


168 


33 


91 


18 


33 


20 


57 


12 


39 


5 


15 


54 


5 


19 


67 


70 


28 


80 


5 


27 


91 


31 


34 


20 


223 


13 


39 


7 


16 


54 


6 


20 


67 


149 


29 


80 


8 


28 


91 


54 


35 


21 


5 


10 


39 


14 


17 


54 


10 


21 


68 


5 


23 


80 


12 


29 


91 


101 


36 


21 


13 


11 


39 


32 


18 


54 


18 


22 


68 


6 


24 


80 


20 


30 


91 


196 


37 


21 


36 


12 


39 


77 


19 


54 


37 


23 


68 


9 


25 


80 


36 


31 


92 


5 


30 


21 


129 


13 


39 


210 


20 


54 


79 


24 


68 


16 


26 


80 


68 


32 


92 


7 


31 


22 


5 


10 


40 


5 


15 


54 


185 


25 


68 


29 


27 


80 


134 


33 


92 


10 


32 


22 


9 


11 


40 


6 


16 


55 


5 


20 


68 


56 


28 


80 


279 


34 


92 


16 


33 


22 


24 


12 


40 


12 


17 


55 


8 


21 


68 


116 


29 


81 


5 


27 


92 


26 


34 


22 


78 


13 


40 


24 


18 


55 


15 


22 


68 


255 


30 


81 


7 


28 


92 


45 


35 


23 


5 


10 


40 


57 


19 


55 


29 


23 


69 


5 


24 


81 


11 


29 


92 


83 


36 


23 


7 


11 


40 


147 


20 


55 


61 


24 


69 


8 


25 


81 


17 


30 


92 


159 


37 


23 


17 


12 


41 


5 


16 


55 


139 


25 


69 


14 


26 


81 


30 


31 


93 


5 


30 


23 


50 


13 


41 


9 


17 


56 


5 


20 


69 


24 


27 


81 


56 


32 


93 


6 


31 


23 


176 


14 


41 


19 


18 


56 


7 


21 


69 


46 


28 


81 


108 


33 


93 


9 


32 


24 


5 


10 


41 


42 


19 


56 


13 


22 


69 


92 


29 


81 


220 


34 


93 


14 


33 


24 


6 


11 


41 


105 


20 


56 


24 


23 


69 


197 


30 


82 


5 


27 


93 


22 


34 


24 


13 


12 


41 


286 


21 


56 


48 


24 


70 


5 


24 


82 


6 


28 


93 


38 


35 


24 


34 


13 


42 


5 


16 


56 


105 


25 


70 


7 


25 


82 


9 


29 


93 


69 


36 


24 


108 


14 


42 


8 


17 


56 


248 


26 


70 


12 


26 


82 


15 


30 


93 


129 


37 


25 


5 


11 


42 


15 


18 


57 


5 


20 


70 


20 


27 


82 


26 


31 


93 


254 


38 


25 


9 


12 


42 


32 


19 


57 


6 


21 


70 


37 


28 


82 


46 


32 


94 


5 


31 


25 


23 


13 


42 


77 


20 


57 


11 


22 


70 


73 


29 


82 


87 


33 


94 


8 


32 


25 


69 


14 


42 


200 


21 


57 


19 


23 


70 


153 


30 


82 


174 


34 


94 


12 


33 


25 


240 


15 


43 


5 


16 


57 


38 


24 


71 


5 


24 


83 


5 


28 


94 


19 


34 


26 


5 


11 


43 


7 


17 


57 


81 


25 


71 


6 


25 


83 


8 


29 


94 


32 


35 


26 


7 


12 


43 


12 


18 


57 


186 


26 


71 


10 


26 


83 


13 


30 


94 


57 


36 


26 


17 


13 


43 


25 


19 


58 


5 


21 


71 


17 


27 


83 


22 


31 


94 


106 


37 


26 


46 


14 


43 


57 


20 


58 


9 


22 


71 


31 


28 


83 


38 


32 


94 


204 


38 


26 


148 


15 


43 


142 


21 


58 


16 


23 


71 


59 


29 


83 


71 


33 


95 


5 


31 


27 


5 


11 


44 


5 


16 


58 


31 


24 


71 


120 


30 


83 


140 


34 


95 


7 


32 


27 


6 


12 


44 


6 


17 


58 


63 


25 


71 


260 


31 


83 


287 


35 


95 


11 


33 


27 


13 


13 


44 


10 


18 


58 


140 


26 


72 


5 


24 


84 


5 


28 


95 


17 


34 


27 


32 


14 


44 


20 


19 


59 


5 


21 


72 


6 


25 


84 


7 


29 


95 


28 


35 


27 


95 


15 


44 


43 


20 


59 


8 


22 


72 


9 


26 


84 


11 


30 


95 


48 


36 


28 


5 


12 


44 


103 


21 


59 


13 


23 


72 


14 


27 


84 


18 


31 


95 


87 


37 


28 


10 


13 


44 


272 


22 


59 


25 


24 


72 


25 


28 


84 


32 


32 


95 


166 


38 


28 


23 


14 


A', 


5 


17 


59 


50 


25 


72 


48 


29 


81 


59 


33 









76 

and 

P^Sio > t | no gene is responsible) (3.49) 

n 2 

= 22 {^(k > 2 ^i' •? ^ ^°' *° or *° + * an< ^ ^° ^ *° or ? o + 1 I no gene)(3.50) 
fcst+i 

P(k > 2 Si Q , k > 2 Si 0+ i\ no gene) (3.51) 

P(2Si = k and 7 7^ «o or »o + 1 I no gene)}. (3.52) 

Eq. (3.47) is less than Eq. 3.51 and the other corresponding terms are equal, therefore 

P(Io is wrong and 2 57 > / | a gene is responsible) 
< PfaSh > t I no gene is responsible) 

and thus, 

P(linkage claim is incorrect) 
< ■P(2'S7 > t | no gene is responsible) 



E 



J2 p (»* > '<-,". I 'o = ') p Co = ') 



all possible results ^ 1=0 
of the stage I 



i=0 



P(the result of stage I) 



3.5 Discussion 

The Monte Carlo results indicated that the relative errors between the probabil- 
ities calculated from formulas and Markov chain simulations were under 7% in the 
dominant cases and under 15% in the recessive cases. Consequently, the approxima- 
tion using the independence assumption for dependent marker loci was acceptable. 

The simulation studies also showed that, in the dominant cases, combining Stage 
I data with Stage II data did not have any significant advantage. The probability 



77 



of allocating the correct marker increased less than by 3%. For the recessive cases, 
there was some advantage when the ASPs were few. The probability of allocating 
the correct marker could increase by as much as 10%. However, it is very diffcult to 
combine data in the theoretical derivation. 

The powers to find the gene depends on the exact gene location between markers. 
Since Table 3.3 and 3.4 were constructed under the least favorable configuration, the 
actual power should be higher. 

The two-stage approach indeed boosted the probability of finding the correct gene 
location under resource constraints. In many cases this probability can increase up to 
20% to 30%. In searching for recessive disease genes, there were several instances when 
the improvement exceeded 35% (Table 3.3). However, if there are enough resources, 
such that almost all available markers can be typed on all ASPs, then the one-stage 
approach may have higher power, (see e.g., N=5000 and APS=10 or N =10,000 and 
ASP=10, 25 in Table 3.3.) However, since the power loss is so small, we may always 
choose the optimal two-stage design. 

As shown in Tables 3.3 and 3.4, it requires much more resources to locate a 
dominant than a recessive disease gene. For example, in a two-stage design with 
25 ASPs and N=1000 we can locate a recessive disease gene with a 0.88 probability 
when £ = 1. But for a dominant disease gene, more than 100 ASPs and/or more than 
N= 10,000 are needed to achieve the same probability. 

Another point worth noting is that phenocopy can severely reduce the probability 
of finding correct gene location. For example, to locate a recessive gene, with resource 
#=5000, 25 ASPs, and e = 1, the chance of finding it is 99.9%. However, when 
6 = 0.5, even if we double the resource to #=10,000 and 50 ASPs, the chance of 
finding the correct locus is only 90%. Althought both of them have about 25 ASPs 
whose disease is caused by gene, those extra phenocopy ASPs reduce the probability 
considerably. 



CHAPTER 4 
TWO-STAGE GENOME SEARCH FOR COMPLEX DISEASE 



In the previous chapter, we have focused on finding a single disease gene. However, 
genetic diseases are not always single-gene diseases. Many of them are complex dis- 
eases, i.e., diseases caused by several genes. For example, insulin-dependent diabetes 
mellitus (IDDM) is influenced by a number of susceptibility genes and environmental 
factors (Luo et al., 1995). A disease phenotype controlled by genes at several dif- 
ferent loci is considered to have "nonallelic heterogeneity" and "when this disease is 
relatively rare and mutation rates are low, individuals within a family are generally 
homogeneous. Locus heterogeneity then leads to the situation that the recombination 
fraction between disease phenotype and marker will be different in different families" 
(Ott, 1991, p. 199). This chapter discusses the probability to find two unlinked reces- 
sive genes that may cause the same disease in a two-stage search under the following 
genetic model. 

4.1 Genetic Model and Assumptions 

Throughout this chapter, we assumed the following: 

• No epistasis, i.e. no interaction between genes. 

• When an individual carries the two disease genes, the penetrance is additive, 
i.e. the probability of being affected for an individual who has two recessive 
genes at two loci is twice as high as for an individual who has two recessive 
genes at only one locus. 



78 



79 



• For an individual who has no disease gene, the probability of being affected is fco 
times the probability of being affected for an individual who has two recessive 
genes at one locus. 

• Receiving one gene has no effect on receiving other genes. 

Other assumptions, necessary for simplifying analytic study are similar to those in 
the previous chapter, are: 

• There are two alleles at each gene locus, denote disease gene d and normal gene 
D. The population frequency of disease gene is p for both loci. 

• Genes are in the middle of two adjacent markers. 

• In this study, only three maps, 5 cM (centimorgan), 10 cM, and 20 cM, are 
available, and every marker are highly polymorphic. Marker's positions on the 
each chromosome are at cM, 5 cM, 10 cM,..., for 5 cM map, cM, 10 cM,..., 
for lOcM map, and cM, 20 cM, 40 cM,..., for 20 cM map. 

• The cost of typing alleles is a constant. 

4.2 Two-stage Genome Search 

For simplicity, we use a slightly different two-stage approach for searching complex 
disease genes. In the first stage, we choose a threshold for statistics instead of choosing 
a number of loci. In the second stage, we choose the loci on different chromosomes 
where the statistics have the highest overall value. Details are as follows: 

Following chapter 3, suppose there are n ASPs and there are enough resources to 
type N markers for ASPs. Again, there are three numbers that must be determined 
in a two-stage design: n\ and m, the number of ASPs and the number of markers to 
be used in Stage I, and k, the threshold of markers to be studied in Stage II. Define 



80 



in the same way as in chapter 3 where i is the index of loci, j is the index of ASP, 
Xij is defined by the assumption 6 in §3.1. If in Stage I, marker i has a iS,- value 
higher than the threshold, k, we will study this marker again in the stage II. Let the 
number of markers that passed Stage I be R. 

In Stage II, R markers on N 2 ASPs are to be typed, where N 2 is the largest number 
subject to the resource constraint and R. Since R is a random variable, N 2 is also a 
random variable. Thus, N 2 is the largest 2, such that mni + Rx < N. We define 

ni+N 2 
J-ni+l 

for stage II. Then, markers that meet the following criteria will be declared having a 
gene nearby: 

• A marker with the uniquely highest 2 S{ is claimed to be the marker nearest to 
the disease gene. 

• A marker group (see page 42), has the uniquely highest score, is claimed to have 
the gene lie between them, 

• If two markers or marker groups have the same highest score but on different 
chromosomes, then declare each having a gene nearby in the corresponding way 
in the above. 

If none of the above applies, then gene location is considered undetermined. 

Once the location(s) has(have) been chosen, same as in the chapter 3, the next 
step is to check whether we can claim linkage. Let t be the 100(1 - a) percentile of the 
maximum of r binomial (n 2 , 0.25) random variables. If 2 5, of the chosen location(s) 
is(are) greater than t, then we claim there is linkage at that location(s). Since we 
imposed some restriction on declaring a maker having a gene nearby, the actual type 
I error will be less than a. The 95% percentile for the unique maximum and marker 
group were given in Table 3.5 and 3.6, respectively. 



81 



4.3 Probability of Allocating the Correct Marker for a Complex Disease 

In this section we discuss an analytical approach and some simulation results. 
§4.3.1 is a preparation for the discussion. 

4.3.1 Possible Parental Genotypes and Trait IBP distribution 

For a disease gene locus, say locus t, let d and D denote the disease and the normal 
gene allele respectively. Let also the population frequency of d be p, and PG denote 
the parental genotype. Let 

Ei = (en,e i2 ) = 

(1,1), if both sibs received two recessive genes at locus i 

(1,0), if sib 1 received two recessive genes and sib 2 did not at locus i 

(0, 1), if sib 2 received two recessive genes and sib 1 did not at locus i 

(0, 0), if none of sibs received two recessive genes at locus i. 

Then the distribution of trait IBD, I t , conditional on parental genotype and £, is 
given in Table 4.1. The numbers in the table is very easy to verify, for example, 
for parental genotype 2, if the order of the chromosomes are specified, say 1, 2, 3, 
and 4 where 1 and 2 belong to father, and 3 and 4 belong to mother, then the 
probability of receiving Dd dd, where D is on the chromosome 4, is p 3 (l - p). If we 
permute the order of chromosomes we will get 4 different permutation, therefore, the 
probability of parents having genotype Dd and dd is 4p 3 (l - p) in a random mating 
population. When parental genotype is Dd and dd, possible sib genotype are Dd 
and dd with equal frequency, therefore, P(£ t =(l, 1)|PG=2)= P(£,=(1,0)|PG=2)= 
P(£=(0,1)|PG=2)=P(£,=(0,0)|PG=2)= }■ In order to illustrate the conditional 
distribution of trait IBD score, subscripts for parental genotype are denoted as d x d 2 
and Dd 3 . When parental genotype is dd and Dd, and E { = (0,0), sib's genotype 



82 

can only be Dd\ or Dd 2 with equal chance, therefore, P(I t =2\PG=2,Ei—(0,0)) — 
P(/ t =l|PG=2,£?,=(0,0))=|. 

4.3.2 Analytic Approach 

Assume there are 2 unlinked genes, namely G\ and G 2 . Without loss generality, 
let Xij and X 2 j be the statistics of the markers next to G\, and X 3 j and X±j be the 
statistics of the markers next to G 2 for the ASP j; /Si, /S2, /S3, and /S4 be the sum 
of Xs respectively in Stage /, as defined at the beginning of this section. Also, let 7\ 
and I 2 be the IBD score of G\ and G 2 for the affected sibs pair. Let the penetrance of 
carrying two recessive gene at one locus is A, two loci is 2A, and none is A; A, where ko 
is the relative risk of being affected for an individual carrying no gene to an individual 
carrying two recessive genes at one locus. The A is unknown but is assumed to be 
small. It will be cancel out in the formula. Then 

P(a gene or both genes are found) 

= P(G 1 is found and G 2 is not) 
+P(G 2 is found and Gi is not) 
+P{G\ and G 2 are found) 

= {P{Gi passes Stage I, G 2 does not pass Stage I, d is found) (4.3) 
+P{Gi and G 2 pass Stage I, G\ is found)} (4.4) 

+{P(G 2 passes Stage I, Gi does not pass I,G 2 is found) (4.5) 

+ P(Gi and G 2 pass Stage I, G 2 is found)} (4.6) 

+P(Gi and G 2 pass Stage I, G\ and G 2 are found) (4.7) 

For a given threshold k, a given event {G\ is found but G 2 is not}, the possible 
relationship between k and 1S1, iS 2 , 1S3, and 1S4 in Stage I, and corresponding 
relationship of 2 Si, 2 S 2 , 2 S 3 , and 2 S 4 in Stage II are given in Table 4.2. For example 
for event 6, given Gi is found but G 2 is not, in Stage I jSi > k, jS 2 > k, 1S3 < k 



83 



s- 

O 

+^ 

u 

> 

B 

>> 

+^> 
O 
Pi 

$ 



a 

ex 

d 

o 

a 

o 



d 

o 

o 

d 

_o 

s 

*Sh 

CO 

Q 
PQ 



H 






5 

Oh 
ST 

o~ 
II 

Oh 


o 


-HlCN -h|CN 
II II 

w w 

o o 

-h|t Oh Oh 


colo> TlOl CNlO) 

II II II 
www 

o o o 

o>g Dh Dh Dh 


-hit -nlro -hit 
II II II 

WWW 

o o o 

^H Dh Oh Oh 


r-»|T -HlCS --l|T 
II II II 

WWW 

o o o 

_ Z- — 2. 


-<|T -h|cn -hIT 
II II II 

WWW 

o o o 

^H Oh Dh Oh 


CM ■—! 

II II 
►«? '"M 

Dh Dh 


CN iH O 
II II II 
l»f < < 

Dh Oh Dh 


CM iH O 
II II II 

•--? •-«? '-H 

Oh Oh Oh 


CM ^h O 
II II II 
•Jf "Jf *Jf 

Oh Oh Oh 


CM -h o 
II II II 

ijf >4f ••? 

Oh Oh Dh 


o 

Oh 

T 1 

o~ 

II 

H 

Dh 


o 


-h]M -h|cn 

A A 

w w 

o" d 

-h|t Dh Dh 

t-i O 
II II 

Dh Oh 


csleo -nleo 
II II 

w w 
d d 

«|S Dh Oh 

^H O 
II II 

^ >»-? 

Dh Dh 


O 


O 


O 


o 

Dh 

T 

Dh 


o 


-h|CS -h|CS 

A A 

w w 

d d 

-1- Oh Oh 

^H O 
II II 

Dh Oh 


cslco -h|C0 

A A 
w w 

d d 

m|2 Oh Oh 

— i o 

II II 
< < 

Oh Oh 


O 


o 


O 


O 
Dh 

i — 1 

¥ 

S. 

oh 


7i 

W 

o" 

rt Dh 

CN 

II 

Dh 


W 

d 

-hIT Oh 

II 

Dh 


W 

d 
-is °* 

CM 

II 

Dh 


O 


o 


o 


S 

0- 


a. 


1 

m 
ft. 
>* 


CN 

I 

i-H 
ft. 


1 

a. 

CM 


CO 

ft. 
1 

1— 1 


T 

ft. 
1 


o 

Dh 




"13 

T3 
Q 


q 

CO 


"T3 

"a 


-a 


q 

Q 

CO 



84 



Table 4.2. Exclusive events for the case "Cn is found but G2 is not." 



event 


Stage I 


Stage II 


iS, 


1S2 


I S3 


IS4 


the largest statistics and relationship 


1 


> 


> 


< 


< 


max( 2 5'i, 2S2) 


2 


> 


< 


< 


< 


2S1 


3 


< 


> 


< 


< 


2S2 


4 


> 


> 


> 


> 


max( 2 5'i, 2-52), and > max(2S , 3 , 2 5 4 ) 


5 


> 


> 


> 


< 


max( 2 S'i, 2^2), and > 2^3 


6 


> 


> 


< 


> 


max( 2 S'i, 2S2), an( i > 2S4 


7 


> 


< 


> 


> 


2S1, and > max( 2 5 , 3, 254) 


8 


< 


> 


> 


> 


2^2, and > max( 2 5'3, 2S4) 


9 


> 


< 


> 


< 


2 S\ and > 2S3 


10 


> 


< 


< 


> 


2Si, and > 2S4 


11 


< 


> 


> 


< 


2S2, and > 2'?3 


12 


< 


> 


< 


> 


2S2, and > 2S4 



and 164 > k, then in the Stage II, the maximum of 251 and 2^2 must have the largest 
value and also larger than 2S4. 

For a given threshold A;, and event {G2 is found but d is not} the possible 
relationship between k and i^i, 1^2, 1S3, and 1S4 in Stage I, and corresponding 
relationship in Stage II are given in Table 4.3. 

For a given threshold k, and event {G\ and G 2 are found} the possible relationship 
between k and iSi, 1S2, 1S3, and 1S4 in Stage I, and corresponding relationship in 
Stage II are given in Table 4.4. 

For a given threshold k, the sum of probabilities (4.3) through (4.7) are 



m-4 33 ./V2 

> 4 / / ^(event i, and / ySj pass Stage I, the largest statistics =h in Stage II ), 

/=0 i=l h=l 



85 



Table 4.3. Exclusive events for the case "G2 is found but Gi is no." 



event 


Stage I 


Stage II 


1S1 


1S2 


iS 3 


IS4 


the largest statistics and relationship 


13 


< 


< 


> 


> 


max( 2 5 , 3, 2S4) 


14 


< 


< 


> 


< 


2-53 


15 


< 


< 


< 


> 


2S4 


16 


> 


> 


> 


> 


max( 2 5 , 3, 2S4), and > max( 2 5'i, 2S2) 


17 


> 


< 


> 


> 


max( 2 S , 3, 2S4), and > 2 5i 


18 


< 


> 


> 


> 


max( 2 S , 3 , 2S4), and > 2 5' 2 


19 


> 


> 


> 


< 


2 53, and > max( 2 5i, 2 5 2 ) 


20 


> 


> 


< 


> 


2S4, and > max( 2 5'i, 252) 


21 


> 


< 


> 


< 


2^3, and > 2 5i 


22 


> 


< 


< 


> 


2^4, and > 2 Si 


23 


< 


> 


> 


< 


2 5 3 , and > 2 S' 2 


24 


< 


> 


< 


> 


2S4, and > 2S2 



Table 4.4. Exclusive events for the case "(?i and G 2 are found." 



event 


Stage I 


Stage II 


1S1 


1 S2 1S3 


1S4 


the largest statistics and relationship 


25 


> 


> > 


> 


max( 2 5 , 3 , 2 5 4 )= max( 2 5'i, ^2) 


26 


> 


< > 


> 


max( 2 5 , 3, 2^4)= 2S1 


27 


< 


> > 


> 


max( 2 5 , 3, 2"S , 4)= 2^2 


28 


> 


> > 


< 


max( 2 S , i, 2^)= 2S3 


29 


> 


> < 


> 


max( 2 .S'i, 2 5 2 )= 2S4 


30 


> 


< > 


< 


2'S'3=2>S'i 


31 


> 


< < 


> 


2 54=2'S'l 


32 


< 


> > 


< 


2«S'3= 2'5'2 


33 


< 


> < 


> 


204=202 



86 

where N 2 = [*=**] M . 

For example for event 25; 

P(event 25, and / iSjS pass Stage I) 

max( 2 5 , 3, 254)= max( 2 5i, 2 S 2 )=h, and all other / 2 Sjs < /1) 
(with assumption of independence) 
= P(\Si > k, 1S2 > k, 1S3 > k, 1S4 > k)P(l \Sj pass Stage I) 
P(max( 2 5 , 3, 2'S'4)= max( 2 5'i, 2 S' 2 )=/i)P(all other / 2 SjS < h), 

where j ^ 1,2,3,4. The distribution of 2 Sj is Binomial(iV 2 , 0.25). Therefore, if we 
want to know probabilities (4.3) through (4.7), we need to know the joint distribution 
of 1S1, 1S2, 1S3 and 1S4 conditional on both sibs are affected. In order to calculate 
this joint distribution, we need to know the joint distribution of X\j and X 2 j and X 3 j 
and X4J conditional on both sibs are affected, which is, 

Pxyuv (4.8) 

= P(X\j = x,X 2 j = y,X3j = u,X4j — v I both affected) (4.9) 

2 2 
= X, 2-/ P^ l i ~ x ' ^2j = y, Xzj = u, X 4 j = v Ii = t'i, / 2 = la | both affected) 

i 1 =0»' 2 =0 

2 2 



= 2_^ 2_^{P(Xij = x,X 2 j = y,X 3j = u,X 4j = v \Ii = z'j, I 2 = t 3 , both affected) 
n =0)2=0 

P(h = i u I 2 = i 2 I both affected)} (4.10) 

2 2 

= E £ P ^ = *• ^ = y I 7 1 = W** = U, X,, = v\I 2 = l 2 ) 
i 1 =0i 2 =0 

P(/i = ii, / 3 = » 2 j both affected) (4.11) 



87 



Table 4.5. Conditional distribution of Xij and X 2 j given /j. 



i 


X 


y 


P{X l} = x, Xij = y\I 1 = i) 


2 



1 

1 





l 
l 


(1 -•?)(!-•» 
(l-tj)t5 


1 



1 


1 





1 
1 


»i(l-»i)(l -*a + *|) 

(1 - *i + *?)*s(l - * 2 ) 

*i(l - *i)* 2 (l - * 3 ) 




j — 




1 



1 






1 
1 


(2*, -*f)(2fa-*l) 

(2« 1 -*J)(l-*a) a 
(1 - ^x) 2 (l - * 2 ) 2 



The probability P(Xij = x, X 2j = y \ h = i) in general case, i.e. gene is 
anywhere in between 2 markers, can be derived from Table 2.3, and is given in Table 
4.5. Where tfj = ^ 2 + (l-^) 2 , j = l,2, B x and 9 2 are the recombination fraction between 
gene and marker 1, and between gene and marker 2, respectively. The probability 
P{X$j = x, X 4 j = y | I 2 = i) is same except using different 0s. 

The probability P(I X = i u I 2 = i 2 | both affected) is equal to, 



P(h = ii, I 2 = ij, both affecte d) 
P(both affected) 

EE E E P{h = H, h = »'a, E u E 2 , PG U PG 2 , both affected) 

_ E x E 2 PGj PG 2 

P(both affected) 
= £E S S P ( both affe cted|/! = i, I 2 = i 2 E u E 2 , PG U PG 2 ) 

£, E 2 PGi PG 2 



88 



P(/, = i u I 2 = %2 ft, ft, PGi, PG* 2 )}/P(both affected) 
EEEE ^(both affected|ft, E 2 )P(I l = i„ h = *2 ft, ft, PGi, PG 2 ) 

£»i £2 PG\ PG2 

P(both affected) 
{££ £ £ ^(both affected|ft, ft)P(/! = in ft, PG 1 )P(/ 2 = ij, ft, PG 2 )} 

Ei £/2 PG\ PG2 

P(both affected) 
{£53££ ^(both affected|ft, ft)P(/i = ii | ft, PGi) 



P(both affected) % ft ^ ^ 

P(ft I PG 1 )P(PG 1 )P(I 2 = i 2 I ft, PG 2 )P(E 2 I PG 2 )P(PG 2 )} 

Because both sibs are affected, hence when &o = the following restriction on sum- 
ming over Ei and E 2 are applied; if ft 7^ (1,1) and ft ^ (1,1) then (ft, ft) 
must be ((1,0), (0,1)) or ((0,1), (1,0)). P(both affected|ft, ft) is given in Table 
4.6. The probabilities P(both affected) = £ i=0 £,=0 P{h — *> ^2 — J, both affected). 
P(/i = i, I 2 = j, both affected) are given as follows. The rest probabilities were given 
in Table 4.1. Therefore, P(h = ij, h = H I both affected) can be found. 

Let Pl = P(PG = 1), p 2 = P{PG = 2), p 3 = P{PG m 3), and p 456 = P{PG = 
4, 5, or 6), Then, 

P(h = 0,I 2 = 0,BA) 

., . 2 /l 1 3 1\ / 9 2 1 

rtx2 (\ 1 3 1\ A 1 3 1 
+2A U P2 2 + ^ 3 3JU P2 2 + T^ P3 3 

„ x2 /1 1 3 r 2 

-HfcA ^ + ^33 

^ 2 / 9 2 1 ^ 

+k ° X (,16 P3 9 + 4 P45 ' 
= ^(2 P2+ , 3 ) 2 



89 



Table 4.6. P(both affected|£q, E 2 ) and possible trait IBD given E\ and E 2 . 



Ex 


E 2 


P(BA\E U E 2 ) 


possible trait IBD score, (h,I 2 ), given E\ and E 2 




(1,1) 


(0,0) 


A 2 


(2,0), (2,1), (2,2) 


(1,1) 


(0,1) 


2A 2 


(2,0), (2,1) 




(1,1) 


(1,0) 


2A 2 


(2,0), (2,1) 




(1,1) 


(1,1) 


4A 2 


(2,2) 




(1,0) 


(0,0) 


KoA 


(1,0),(1,1),(1,2),(0,0),(0,1),(0,2) 




(1,0) 


(0,1) 


A 2 


(1,0),(1,1),(0,0),(0,1) 




(1,0) 


(1,0) 


ZkqA 


(1,0),(1,1),(0,0),(0,1) 




(1,0) 


(1,1) 


2A 2 


(1,2),(0,2) 




(0,1) 


(0,0) 


k X 2 


(1,0),(1,1),(1,2),(0,0),(0,1),(0,2) 




(0,1) 


(0,1) 


ZKqA 


(1,0),(1,1),(0,0),(0,1) 




(0,1) 


(1,0) 


A 2 


(1,0),(1,1),(0,0),(0,1) 




(0,1) 


(1,1) 


2A 2 


(1,2),(0,2) 




(0,0) 


(0,0) 


k X 


(2,0),(2,1),(2,2),(1,0),(1,1),(1,2),(0,0),(0,1),(0,2) 




(0,0) 


(0,1) 


koX 


(2,0),(2,1),(1,0),(1,1),(0,0),(0,1) 




(0,0) 


(1,0) 


KoA 


(2,0),(2,1),(1,0),(1,1),(0,0),(0,1) 




(0,0) 


(1,1) 


A 2 


(2,2),(1,2),(0,2) 















90 



+£j;M 2 (2p 2 + p 3 )(2p 2 + 3p 3 + 4p 456 ) 



+ ^o 2 A 2 (P3 + 2 j045 6) 2 



P(h = 1, l 2 = 0, BA) = P(h = 2, I 2 = 1, flA) 

ot _ .j/1 1 3 2\ / 9 2 1 \ 
2koX U P2 2 + T6 P3 3 ) U P3 9 + i P45 V 

^ Q . 2 /l 1 3 2\ /l 1 3 1\ 
+2A U*2 + 16*3^2 + 16*3 J 

^ tff 1 1 3 2\ /l 1 3 1\ 

+4M U P2 2 + ^ 3 3jU P2 2 + ^ P3 3j 

+2M U P2 2 + I^ 3 9 + 2 P ™ P ™) U P2 2 + l6 P3 3 J 

,. 3 v2/l 1 9 4 1 \ /9 2 1 \ 2 

+k ° X U P2 2 + T6 P3 9 + 2 P456P456 J U P3 9 + i P45 V 

^A 2 (p 2 +p 3 )(2p2+P3) 

+Y^k Q x 2 (p 2 + p 3 )(p 2 + p 3 + p 456 ) 

+ ^oA 2 (p 2 + 2p 3 + 4p 4 5 6 )(2p 2 + p 3 ) 
+-^ k o x2 (P2 + 2p 3 + 4p 456 )(p 3 + 2p 456 ) 

P(h = 2J 2 = 0, BA) = P(h = 0J 2 = 2, BA) 
A2 ( Pl + \ P2 + Te*) (l6 P 4 + \ P ™) 
+4X2 ( Pl + l P > + h P3 )(W2 + h P3 l) 



91 



119 2 1 



9 2 1 



( ^ + 16 P3 9 + 4 P456 > I ^ P3 « H ' 4 P456 



119 3 1 



+2fc A 7P2 7T + TTP3 77 + TP456 TP2^ + TTP3X 



V 2 16" 9 4 



16 r 9 4' 



113 1 



4 r 2 16 r 3 



128 



A 2 (16pi + 4p 2 + Pa)(4p 2 + 3p 3 + 2p 4 5e) 



128 



k Q \ 2 (4p2 + 3p 3 + 4p 4 56)(2p 2 + Pa) 



1 



+—k 2 \ 2 {2p 2 + 3p 3 + 4p 456 )(P3 + 2 P 45 6 ) 



P{h = l,I 2 = l,BA) 



11 3 2\/l 1 9 4 1 



l 4 P2 2 + l6 P3 3JU P2 2 + 16 P3 9 + 2 P456 



.11 3 2 
U P2 2 + 16 P3 3 



,11 3 2 
4M [-p, i + ^3 



.11 9 4 1 
"*° A U P2 2 + T6 P3 9 + 2 P456 



^A 2 (p 2 +p 3 ) 2 

+— fcoA 2 (p 2 + Pa)(2p 2 + 3p 3 + 4p 4 5e) 
16 

+ —/c 2 A 2 (p 2 + 2p3+4p456) 2 

o4 






P(/, = 2, / 2 = 1, BA) = P{h = l,/i = 2, SA) 

,./ 1 1 \ /l 1 9 4 1 \ 

A \ Pl + A P2 + 16 P3 J U P2 2 + l6 P3 9 + 2 P456 J 



92 



, ,2x2 /I 1 9 3 1 \ /l 1 9 4 1 

+M U P2 2 + l6 P3 9 + 4 P456 J U P2 2 + T^ P3 9 + 2 P456 

^ou \*( l l 9 3 1 \ A 1 3 2\ 
+2M [- aP2 ~ + -pa- + ^ 456 J (jft- + -Pag J 

— A 2 (16p! + 4p 2 + p 3 )(5p 2 + 7p 3 + 4p 456 ) 

+ 64^° A ^ 2p2 + 3p3 + 4 P 456 )(^2 + Pa) 

+ 128 k2 ° X ^ 2p2 + 3P3 + 4 ^56)(P2 + 2p 3 + 4p 4 56) 

P(I 1 = 2 t I 2 = 2,BA) 

ox2/^ , 1 1 \ /l 1 9 4 1 \ 

= 2A ^ + ^ + - P3 j ^ P2 - + _,3_ + _ p456 j 

/ 1 1 \ 2 

+4A 2 U + -pa + — P3 



, ;2 , 2 /l 1 9 3 1 \ 
+k ° X {4 P2 2 + T6 P3 9 + 4 P456 ) 



1 
128 

1 



256 



A 2 (16p! + 4p 2 + p 3 ) (32p! + 10p 2 + 5p 3 + 4jo 4 56) 
& 2 A 2 (2/> 2 + 3p 3 + 4p 456 ) 2 



Therefore, the joint distribution of X u and X 2j and X 3j and X^- conditional on 
both sibs are affected, p xyuv , can be found. Even though p xyuv is known, it takes too 
much time to compute the exact joint distribution of ,$i, ,S 2 , iS 3 and jS 4 conditional 
on both sibs are affected using the same method we used in the chapter 3 for joint 



93 



distribution of ,5 and jSi- Therefore, Monte Carlo integration were used to find the 
conditional joint distribution of iS\, jSa, 1S3 and 1S4. 

4.3.3 Simulation under More Realistic Assumptions 

For the same reason as in the simple disease case, we also performed simulations 
for an independent model and a Markov chain model without combining Stage I and 
Stage II data. The total length of chromosomes is 3300 cM and divided into 22 
chromosomes. The number of genes is two. The combinations of parameters are: 
resource N = 2000, 5000, 7500, 10,000, and 20,000; number of ASP n = 50, 100, and 
150; population frequency (P.F.) of disease gene = 0.01, 0.2, and 0.5; relative risk, 
ko, of being affected of carrying no gene vs. carrying two recessive genes at one locus 
is equal to 0, 0.05, 0.1 and 1.0. Map densities used are 5 cM, 10 cM, and 20 cM. 
Threshold k from 1 to n 1? and n x from 5 to n - 5 with increment 5. The simulations 
were conducted similar to the simulation in chapter 3. We simulated 22 pairs of 
chromosomes instead of one; two genes instead of one. The details of simulation are 
as follows. 

1. Reading in parameters, resource N, P.F., A; , number of ASP n, and map, n x , 
and k of the design. 

2. Generate gene location from uniform distribution(0, 3300/22) twice, first one for 
the gene on chromosome 1, and second one for the gene on chromosome 2. Then 
follow the rest of step 2 on page 61. For those chromosomes were not chosen, 
set /=0. Generate joint trait IBD for two loci. 

3. Follow step 3 and 4 on page 61 for every chromosome. 

4. Repeat step 2 and 3 ni times and then calculated iS{. 

5. Check if any of i5i, iS 2 , 1S3, and 1 5 4 > k. If yes, then calculate N 2 according 
to the resource constraint, choose markers whose 2 Si > k for Stage II study, 



94 



and go to the next step. If not, record the detection as a failure and go back to 
step 2. 

6. The XijB in Stage II are generated the same way as those in Stage I, except 
only the chosen markers are used and repeat A^ times instead of n^ 

7. Check whether 251 or 2 5a, or 2 S 3 or 2^4, according to which one passed Stage 
I, are the largest among all 2 S';s in Stage II. If any one of them is the largest, it 
is a success; otherwise it is a failure. Go back to step 2 until enough simulation 
has been done. 

The programs were compiled and ran in the same computing environment men- 
tioned in chapter 3. 

4.3.4 Results 

Based on analytic computation and simulation, the design with the highest prob- 
ability of finding the correct marker(s) was identified for a disease with two unlinked 
recessive genes with additive penetrance. These optimal designs are given in Ta- 
ble 4.7. 

In the tables, the P.F. column shows the population frequency of the disease 
gene; the k columns, the relative risk; the map columns, the map density; the k 
column, the threshold for passing Stage I for Stage II study; the m columns, the 
number of ASP used in Stage I; the F2 column, the probability of locating the right 
marker by the best two-stage design obtained by analytic formula; the 1.1, 1.2, and 1.3 
columns show the probabilities, obtained by simulation with independent assumption, 
of finding the markers next to the genes, within two intervals, and within three 
intervals, respectively, by the two-stage design. M.l, M.2, and M.3 columns show the 
probabilities corresponding to 1.1, 1.2, and 1.3 obtained by simulation with Markov 
chain assumption without combining first- and second-stage data; The Fl column 



95 



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110 



gives the that probability of the best one-stage design, i.e., with optimal m and n 
subject to mn < N obtained by analytic formula. The F2— Fl column shows the 
increase in probability of two-stage design over one-stage. 

4.4 Discussion 

The Monte Carlo results indicated that the relative errors between the probabil- 
ities calculated from formulas and Markov chain simulations were almost negligible 
when using a 20 cM map; little discrepancy for 10 cM; a greater difference for 5 cM 
maps. This is most likely due to the fact that under a Markov chain model, when 
using a denser map, we need more samples to separate out the markers next to the 
markers adjacent to the genes. If we examine 1.2, 1.3, M.2, and M.3 columns. We will 
find that for a denser map it can still narrow the gene location to an area wider than 
the interval between two adjacent markers. For example, for N=10,000, ASP=50, 
P.F.=0.01, and Jfc = 0, M.l is only 0.88 but M.2 is 0.97. This shows that if a denser 
map is used, we might need more ASPs to pinpoint the gene location. 

The population frequency of the disease genes and relative risk fco of carrying no 
gene verses carrying two recessive genes at one locus have interaction on the proba- 
bility of finding correct markers. For given resource, ASP, and population frequency, 
the probability of finding correct markers is a decreasing function of relative risk k . 
If population frequency is increased, then the speed of the decrease in probability in 
relative risk will be slower. This is a very intuitive result; when population frequency 
of disease is low and relative risk of to be affected for carrying no gene vs. carrying 
two recessive gene at one locus is high, there will be a lots of phenocopy cases in 
the sample. Hence, the probability to find correct markers will be small. On the 
other hand, when population frequency of disease is high, and the relative risk is low, 
there will only be a small number of phenocopy cases in the sample. Therefore, the 
probability to find correct markers will be large. 



Ill 



Two-stage approach once again provides big improvement in many cases, for ex- 
ample, when N=2,000, ASP=100, population frequency of disease gene=0.5, and the 
relative risk = 0.1, the probability of finding at least one of two disease gene loci is 
boosted from 30% to 80% by a two-stage design. The improvement over 40% were 
marked with * in Table 4.7. Also once again, two-stage approach performs no better 
or worse than one-stage approach is observed, all the worse cases happened when all 
the ASP can be typed in one-stage designs. 



CHAPTER 5 
CONCLUDING REMARKS 



Gene hunting is a difficult task, it usually requires a vast amount of resources. 
Two-stage approach is one way to reduce cost. This dissertation gives the optimal 
designs that can utilize resources very efficiently. 

When a researcher wants to use the results in this dissertation, with information 
from other researchs or other consideration, he/she must postulate if the disease is 
caused by a single gene or multiple genes, the percentage of phenocopy, population 
frequency of disease gene, and relative risk. If funding and sample (ASPs) were 
already obtained, researcher should check the tables to find the combination of, the 
genetic model (dominant or recessive), funding condition (N), size of the sample (n), 
and heterogeneity (or population frequency and relative risk) that match his/her 
situation, and then use the design in the tables. If the funding and sample are not 
yet obtained, the researcher should consult with the tables and find the design that 
would fit his/her possible funding and sample conditions. For example, if a researcher 
believes that the disease is caused by a dominant gene, and think that 100 ASP will 
be available for typing and can only inquire funding enough to type 5,000 makers 
loci for ASPs, then using 125 markers and 20 ASPs in Stage I, choose 17 markers 
for Stage II study would be a design that provides a approximate 75% probability of 
finding the right location. On the other hand, if he/she can not collect more than 75 
ASPs for the study, and funding for typing 10,000 markers is available, then use 275 
markers and 15 ASPs in Stage I and choose 59 markers for Stage II would also provide 
a approximate 75% probability of finding the right location. If a 75% probability is 
not good enough, then the researcher has to find extra funding or more ASP for the 



112 



113 



study, there is no other two-stage design can do better than the design obtained from 
the tables. If the combination of the parameters is out of the range of the tables given 
in this dissertation, then researcher should obtain the programs from this author to 
calculate the best design. 

In the future, extending the method presented in this dissertation to searching 
multiple complex disease genes (more than two), a simple quantitative trait loci 
(QTL) and complex QTL's, and to incorporate Risch's risk ratio method are planned. 
Extension to incorporate different genetic model should also be included. 






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Inc. 



BIOGRAPHICAL SKETCH 

Chi-Hse Teng was born and grew up in Taipei, Taiwan, Republic of China. He re- 
ceived his B.S. in mathematics from National Taiwan University in 1987. He servered 
as a military policeman in the Chinese Army from 1987 to 1989. After an honorable 
discharge, was a research assistant at the Institute of Statistics, Academia Sinica, 
R.O.C., before beginning his graduate study in the Department of Statistics at the 
University of Florida. He won a prize in the student submission competition of the 
Third Annual Florida Epidemiology Meeting in July 1997. 



120 



I certify that I have read this study and that in my opinion it conforms to accept- 
able standards of scholarly presentation and is fully adequate, in scope and quality, 
as a dissertation for the degree of Doctor of Philosophy. 



S%U~ 



C~ ! 



Mark C.-K. Yan 
Professor of St 




I certify that I have read this study and that in my opinion it conforms to accept- 
able standards of scholarly presentation and is fully adequate, in scope and quality, 
as a dissertation for the degree of Doctor of Philosophy. 




Randolph L. Carter 
Professor of Statistics 



/ LmA^o 



I certify that I have read this study and that in my opinion it conforms to accept- 
able standards of scholarly presentation and is fully adequate, in scope and quality, 
as a dissertation for the degree of Doctor of Philosophy. 



?W^ Q-Yhutz. 



Frank G. Martin 
Professor of Statistics 



I certify that I have read this study and that in my opinion it conforms to accept- 
able standards of scholarly presentation and is fully adequate, in scope and quality, 
as a dissertation for the degree of Doctor of Philosophy. 

feBlf\ ft W<-£>U*roy 



Susan P. McGorrary 

Research Assistant Professor of Statistics 



I certify that I have read this study and that in my opinion it conforms to accept- 
able standards of scholarly presentation and is fully adequate, in scope and quality, 
as a dissertation for the degree of Doctor of Philosophy. 



^d^pO 



&- 



Jin Xiong She 
Associate Professor of Pathology and 
Laboratory Medicine 



This dissertation was submitted to the Graduate Faculty of the Department of 
Statistics in the College of Liberal Arts and Sciences and to the Graduate School and 
was accepted as partial fulfillment of the requirements for the degree of Doctor of 
Philosophy. 

December, 1997 



Dean, Graduate School 









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nuivFRSITY OF FLORIDA 

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